Physics

A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

a)Find the speed at which the ball clears the wall.

I calculated this correctly and got 18.13 m/s.

(b) Find the vertical distance by which the ball clears the wall.

I know that dy=vy0*t but I tried using that equation and couldn't get the answer.

I did this and it was incorrect
18.13*sin53*2.2=dy
32.85-6.7=25.45

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.

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  1. On part b,
    You need to find how long it takes the ball to go th 24.0m in the x direction. Use that time for the y calculation.

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  2. Sorry, I see that was given.

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  3. Yeah :( thanks for trying though, I tried using the v0sin53 and got 18.13*sin53= 14.48

    then I tried plugging that into the y, to get y=14.48sin53(2.2)-1/2(-9.8)(2.2^2)
    I got 25.44-23.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong

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  4. I calculated the ball was thrown at a speed of 18.12m/s.

    Vx is 24m/2.2s=10.91m/s
    Vx=V(cos(53))
    That gives V as 18.12m/s

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  5. I get 14.47m/s for Vy.

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  6. Using that...
    Y=14.47m/s(2.2s)-(1/2)(9.8m/s)(2.2)^2
    Y=31.834m - 23.716m
    =8.118m
    So it clears the wall by 8.118m - 6.7m
    =1.418m

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