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A rock is thrown upwards from the ground with initial velocity V_0 at same instant is dropped from rest a distance y_0 above the ground directly above the first roc...

Question

A rock is thrown upwards from the ground with initial velocity V_0 at same instant is dropped from rest a distance y_0 above the ground directly above the first rock find the time when the two rocks collide in terms of the given quantities and acceleration of gravity g. find height from the 2nd rock must be dropped so that two rocks collide at the highest point attained by the 1st rock

a rock is thrown upwards from the ground with initial velocity V_0 at same instant is dropped from rest a distance y_0 above the ground directly above the first rock find the time when the two rocks collide in terms of the given quantities and acceleration of gravity g. find height from the 2nd rock must be dropped so that two rocks collide at the highest point attained by the 1st rock



Answers

A ball is thrown straight up from the ground with speed $v_0$. At the same instant, a second ball is dropped from rest from a height $H$, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of $H$ in terms of $v_0$ and $g$ such that at the instant when the balls collide, the first ball is at the highest point of its motion.

For this question are asked to consider two bolts. One ball is dropped straight downwards from a height H and then another. Balls thrown straight upwards from a initial height of zero with an initial velocity B not were asked to find a time in which they collide well, the height in which they collide. We can call that age subsea, and for the first ball that's gonna be equal to the height in which it was dropped. H Plus, there's no initial velocity. So, uh, this V not times the time T term is zero because there's no initial velocity. It was starting at rest, plus 1/2 times acceleration due to drop gravity g times the time T squared and for the second ball, the time in which they collide or the position which they collide. H subsea. It's the same as the first ball, but it's equation is vey not times the time t um, because it starts out with an initial velocity, but at H value of zero so that value isn't included, plus 1/2 times gravity times T square, so we can set these two equations equal to each other and salt for tea. So what we have is a tch. Plus 1/2 GT squared is equal to V not times the Time t plus 1/2 GT squared. Well, obviously the 1/2 GT squared cancel. You couldn't just subtract those out of both sides of the equation, so that's gone. And that's gone. We can solve for Time T when we find that tea is simply equal to the height in which the first ball was dropped, divided by the initial velocity of the other girl. When we come box set in as their solution for a part, B wants us to determine an expression for the height in terms of the initial velocity and gravity. So for Part B, we know that the final velocity of the ball here in this situation because we're considering when the first we're considering that this collision is gonna happen at the highest point of the motion of the first ball. So at the highest point of the motion of the first ball, the final velocity squared is equal to zero. But the equation for velocity is vey not squared. Um, plus two times gravity times the height H. So now putting H selling for H here we find that H is equal to V not squared over to G. So we did what we were asked to do. We found H in terms of V not in gravity, g. So we can box that in Is their solution to part B. That's the final part of our question.

How long we have given a stone is launched straight up by a selling sort. It's initially speedy's 19.6 m per second, and the stone is 1.50 m above the ground. Suppose this is ground and this is 1.50 m. And suppose this will go like this and hit the ground in party. We have to find how high above the ground does the stone rise. So we have to find stuff. All we know that this distance is equal to 1.50 m. Now we will find this edge, so this edge will be called. We know that formula is it called We scared by two G. V is 19.6 square two and 29.80 so this match will be called to 19.6 square, divided by two and 29.80 So this is a call to 19.6 meter, so the maximum height from the ground will be equal to 1.50 plus 19.6. So this is a cold 1.50 plus 19.6. That means 21 0.1 meter this is the maximum distance of the ground. Mm. Now, in part, we we have to find how much time lapses before the stone hits the ground. So in this problem, we will use as is equal to you be blessed half a T squared. So the display vertical displacement of this Estonia's minus 1.50 and US 19.6 m pass again in 30. And we know that value of exhalation is minus 9.80 in the downward direction. So here this will be called 4.9 in 23 square. So now we will solve this quadratic equation. 4.9 T square minus 19.6 in 30 minus 1.50 cal 20 So this time is equal, though four point one second. This is the answer for part. We thank you.

So we're trying to find how far Stone goes into the air, given the initial velocity of 19.6, and we can figure out the final velocity zero because if you think about it, it's like a shooting up into the sky. In order for it to come down, it has to stop just for a sec. And we can view that split second where it stops as the final velocity being zero. And so we have acceleration, final initial velocity and displacement that we need to Seoul for. So using the equation final velocity squared, my initial velocity squared is equal. Teoh you to a times. The displacement of why I don't velocity, as we said, is zero. The initial velocity is 19.6 square, and that is equal to two times negative. 9.8, which is a negative 19.6. Have this Wissman and solving from this school give displacement of why is equal tonight. No, it's not 19.6, as it tells us in the problem that stone his shot already 1.5 meters above the ground, which would say that the actual displacement is 21.1 as 19.6 plus 15 is 21 point, and that's how high above the ground some will rise before it comes back down to the ground. Now, to find how much time it takes in stone hit the ground, we can use our equation that displacement Why equals G hi times displacement T plus 0.5 acceleration times Displacement t squared. And now we know that the initial velocity is given its 19.6. The displacement of tea is what we're trying to solve for, and that's going to be minus 4.9 as gravity is our acceleration times, displacement of T squared and this of Eagle Teoh Negative 1.2. Then if we're solving from T, this is gonna be another problem where we have to set it up. Like quadratic has just noticed that if you view the displacement of T like X, we have an X where a B to see and this would give that the displacement of tea is equal to the square. 19.6 plus or minus. The square root of 19.6 squared minus four times 4.9 times negative, 1.2 over 9.8 solving. From here, we ignore the negative one. Um, it's 19.6 plus, as if we did 19.6 minus would have a negative amount of time and you can never have a negative amount of time. So solving this out, we'll get the displacement of tea is equal. Teoh for 0.6 seconds.

Party of this problem is asking what the total elapsed time of the slower rock will be on its entire journey. Before we're able to calculate that, we need to figure out what the initial velocity of the slow Iraq is on longest journey. And so were given this information Accelerations name not one ate the initial position. Zero The initial speed of the initial velocity of the Slow Rock is just the initial lawsuit that slow rock and then of the large rockets three times slow rock. And then the time elapsed for the fast Arrakis. Ten seconds. So since the time elapsed for the toll trip is ten seconds, we know that it takes five seconds for the Fast rock to reach peak height. And then once it does reach peak, I our velocity will be zero. And this is the velocity of To which I call the Faster Rock. Now, in order to figure out the initial velocity, I'm going to use this can Mack Equation musical Beano plus extortion times time via the peak zero. Our initial velocity is this is off the fast rocked by the way, So be not to which is three v not one plus the acceleration. Negative nine twenty. Time is the time and this is the time from the start. Until it reaches peak height, which is five seconds. We can solve this For the initial velocity of the Slow Rock, we get sixteen point three three meters per second. Now that we have this, we can figure out the total time for the smaller rock. So if ve easier one is equal to sixteen point three eighty three years for second we can use the same can Matic equation is equal to be not plus eighty. The difference is before I applied it to the Faster Rock in order to find this But now I'm going to apply to the Slow Rock. Now that we know this and because we're playing too slow Rock, I gotta remember to put that sub script there, The one there. And so we're just going to go up to the peak. So the final velocity zero, this is sixteen point three three acceleration. Still naive, non pointy. And then tea Salling, we get tea is equal to one point six seven seconds. That is just the time it takes to go from the beginning until the top of the height there. So we have to double this to get the total time elapsed. So T total of the slow one number one is double that, or three point three three seconds. And that's the answer to party. We also need to figure out the relative heights of thes travel. And so that is party. So to do this, I'm going to figure out the total height that the faster rock travels. And I'm gonna figure out the total height that the slower rock travels and I'm going to divide them. Figure out how the ratio of the two acts, which will give me the final answer. So re not to is three b No. One and we calculated the not one before ISS sixteen point three three meters per second. So this is being forty eight point nine nine years for second again, I'm going to let the final glossy B zero we're just going to go to the apex. And then the time alas, for this is given in the problem is five seconds. And so I'm going to use the Kinnah Matic equation that relates these quantities. Next, my sex and I equals one half and I'm gonna plug in my glossy. Really, It's a difference of the final in the initial, but in this case, the initial velocity zero. So I just have the final velocity of forty eight point nine nine times the time last, which is five seconds. This gives distance traveled of one hundred twenty two point five meters. I'm going to do the same thing for the slow Rock. Use the same equation X minus X is equal to one half. In this case, the initial velocity is sixteen because it's Laura and then the time elapsed until it reaches the apex is one point six seven seconds. So do me this hour I get next. My sex nine is equal to thirteen point sixty four meters. So really, they call this H and I want to know what this is in terms of h. So to do that, we divide the two. I'm going to note the distance of the faster rock by putting it to there, and it's one hundred and twenty two point five meters. But I want to divide. But with the slow rockets going to which is thirteen point sixty four I'm gonna call that age in the unit, sir, doing the cell I get. That's approximately nine H Self H is the height that the slow rot goes than the high, the faster Rocco's nine Times Age. So nine times is high, and that's the final answer of Part B.


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