You stand 17.5 m from a wall holding a softball. You throw the softball at the...

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You stand 17.5 m from a wall holding a softball. You throw the softball at the...

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 32.5∘ from the ground with an initial speed of 30.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

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Answer 1

Answer #1

Time taken by the ball to reach the wall can be calculated by applying equation of motion in x-direction

s = u_x*t

u_x = 30.5 Cos32.5^o = 25.72 m/s

u_y = 30.5 Sin32.5^o = 16.39 m/s

17.5 = 25.72*t

t = 0.68 s

Now, we have to calculate the height of the softball at t = 0.68 s using 2nd equation of motion in vertical direction,

s = u_yt + (1/2)at²

h = 16.39*0.68 + (1/2)*(-9.8)*(0.68)²

h = 8.88 m

Therefore, height of the softball at the point of impact on the wall is 8.88m

I hope i was able to answer your question. If you need anymore explanation, please ask in comments. I would be happy to help.

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Answer 2

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