For this problem on the topic of motion. In two dimensions, We are told that a rock is thrown from a 15 m building with a speed of 30 m/s at an angle of 33° above the horizontal. We want to find the maximum height above the roof that the rock will reach its speed just before it strikes the ground. The horizontal range from the base of the building that the rock will strike the ground. And we want to draw graphs for X versus T. Why versus T. V, X versus T. And B. Y versus T. Now we'll take the region of coordinates of the roof and let the positive Y. Direction be upward and the positive X. Direction be to the right. The rock moves in projectile motion with a X equal to zero and a Y equal to minus G. We can then set up constant acceleration equation for the X and Y components of the motion. Firstly, the initial velocity has X and Y components. We not cosign Alpha not and we not sign Alpha not, respectively, which is 25.2 m per 2nd and 16.3 m per second respectively. So firstly for part A to find the maximum height, we know that at this point V Y Must equal to zero. And we can use the equation. The y squared is equal to Vinod Y squared less to a Y into Y minus Why not? So we can rearrange this equation and solve for its displacement at the maximum height. Why minus? Why not? Is the speed at the maximum height B Y squared? Mind its initial why component of its speed, we're not Y squared Over two times the vertical acceleration to AY. And so this is zero at the maximum height minus V, not Y 16.3 m per second squared, Divided by two times -9.8 m/km2, which is G. And so calculating, we get the maximum height that the stone will reach of Iraq will reach other 13.6 m for part B. We want to find the speed of the rock just before it strikes the ground. Now we'll find the velocity by solving for X and Y components and so it's X component would remain the same. The X is equal to the not X. This is not affected by gravity and so the ball strikes the ground at a X. Component of speed 25.2 m per second since the acceleration in the X direction A X. Is equal to zero. Now we need to find the Y component of the velocity. So we'll use the equation. The Y squared is equal to we not Y squared plus to a Y. Into why minus why not? And so we get V. Y. To be minus the square root of the not peanut, Y squared plus to a Y into y minus. Why not? We take the negative route because the at the ground the rock is traveling downward and so this is minus the square root of the initial speed. 16 three m/s squared plus To into -9.8 meters a square second multiplied by the vertical displacement minus 15 m. This gives us the why component of the speed as the balls or the rock strikes the ground to b minus 23 .7 m/s. And so therefore the magnitude of this velocity V is the square root of the sum of squares of the X and Y components, the X squared plus b, Y squared. This is the square root of 25.2 m/s squared plus minus 23.7 m per second squared, Which gives us the final velocity of the rock to be 34 .6 m/s. Next, for part, see we want to find the horizontal range from the base of the building to the point where the rock strikes the ground. Now we'll use the vertical motion, the Y component to find the time that the rock spends in the air. And we know that this time T. Is equal to the final speed V. Y minus the initial speed in the Y direction. Do not why? Over the acceleration A. Y. And this is minus 23 .7 m/s minus its initial speed. 16.3 m per second upward of uh minus G, which is 9.8 m/km2. This gives us your time Of 4.08 seconds. Now we can use this time to calculate the horizontal range so we can use another equation, X minus x not is equal to B, not X times T bless a half. Hey, X t squared. So this is the initial velocity in the X direction 25.2 m/s times a time of 4.08 seconds as the second term, which is zero. Since the acceleration in the X direction is zero, This gives us the range of the rock to be 100 and three meters. And lastly, we want to draw graphs for X versus T, Y versus T. V, X versus T, and V Y versus T, which look as follows. And here we have the graph of X versus t, why versus T. The X component of velocity VX versus T. And the Y component of velocity V y versus T.