Aball is thrown horizontally from the top of
a building 130 m high. the ball strikes the
ground 64 m horizontally from the point of
release.
what is the speed of the ball just before it
strikes the ground?
answer in units of m/s.

answer:

a) the ball falls 43.7m from inital vertical speed u = 0 accelerating at a= 9.8m/s^2

s = ut +1/2 x a x t^2

t = sqrt(2s/a) = sqrt(2 x 43.7 / 9.8) = 2.99s

b) ball travels 114m in 2.99s before hitting the ground, speed = distance/time

speed = 114/2.99 = 38.12m/s

c) the ball has no horizontal forces acting on it (assuming no air resistance) so it remains at constant speed, therefore final horizontal speed is the same as the initial speed = 38.12m/s

d) the final vertical speed is given by:

v = u +at, where u=0, a = 9.8m/s^2, t = 2.99s

v = 9.8 x 2.99 = 29.3m/s

explanation:

hope that !