please .. A boy reaches out of a window and tosses a ball straight up with a spe

Question

please ..

A boy reaches out of a window and tosses a ball straight up with a speed of 11 m/s. The ball is 21 m above the ground as he releases it. Use energy to find the ball's maximum height above the ground. Express your answer to two significant figures and include the appropriate units. Use energy to find the ball's speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. Use energy to find the speed of impact on the ground.

Explanation / Answer

(a) His maximum height is when all the speed is turned into height (Ekinetic --> Epotential)
Ek = Ep
½mv² = mgh (divide bij m)
½v² = gh (multiply by 2)
v² = 2gh (insert v and g)
11² = 2*9,81*h
121 = 19,62h (divide by 19.62)
6.16 = h

So the added height is 6.16m. Add this to the 16m at which it started: 21 + 6.16 = 27.16m

(b) Then it changes his potential energy back to kinetic energy.
Ep = Ek
mgh = ½mv² (divide by m)
gh = ½v² (multipy by 2)
2gh = v² (insert g and h. For h use the height above the window: 7.34m)
2*9.81*6.16 = v²
120.85 = v²
10.99 = v

So it seems it passes down with the same speed it went up.

(c) Now use the same way as in b, but change the height from 6.16 to 27.16
Ep = Ek
mgh = ½mv² (divide by m)
gh = ½v² (multiply by 2)
2gh = v² (insert g and h)
2 * 9.81 * 27.16 = v²
532.87 = v² (sqrt)
23.08 = v

So the speed of impact on the ground is 2308 m/s.

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