8Iriend is standing on top of a building 50 m above you, and for whatever reason

Question

8Iriend is standing on top of a building 50 m above you, and for whatever reason you wish to toss a baseball up to them. (a) How fast must you throw the ball so that it just barely reaches your friend? (b) You realize that you don't have a good quantitative feel for how fast you're going to throw the ball, so instead you just throw it as hard as you can. The ball reaches 90 mph just as it leaves your hand. How fast will your friend have to react in order to catch the ball on its way up? (c) You forgot that your friend is generally not the athletic type. They don't react fast enough to catch the ball on the way up, and they fail to catch it on its way down. How much time will you have to jump out of the way of the ball after your friend misses their second chance to catch it?

Explanation / Answer

5.

a)

h = height = 50 m

v = speed of launch = ?

m = mass of the ball

using conservation of energy

kinetic energy at the bottom = potential energy at the top

(0.5) m v2 = mgh

v = sqrt(2gh) = sqrt(2 x 9.8 x 50) = 31.31 m/s

b)

vo = initial speed = 90 mph = 40.23 m/s

t = time to react

Y = displacement = height = 50 m

a = acceleration = - 9.8 m/s2

using the equation

Y = Vo t + (0.5) a t2

50 = (40.23) t + (0.5) (-9.8) t2

t = 1.53 sec

c)

vo = initial speed = 90 mph = 40.23 m/s

t = time to move out

Y = displacement = 0 m

a = acceleration = - 9.8 m/s2

using the equation

Y = Vo t + (0.5) a t2

0 = (40.23) t + (0.5) (-9.8) t2

t = 8.21 sec

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