A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 15.0 m/s. Assuming negligible air resistance, what is the speed of the ball just before hitting the ground?

36.4m/s

Explanation:

According to equation of motion,

v² = u²+2as where;

v is the final velocity

u is the initial velocity = 15m/s

a is the acceleration due to gravity = + g (it is positive because the body is thrown downwards.)

g = 10m/s²

s will be the distance from the top of the tower from which it is thrown

S = 55.0m

Substituting this value to get the final velocity, we have;

v² = 15²+2 (10) (55)

v² = 225+1100

v² = 1325

v = √1325

v = 36.4m/s

Therefore, the speed of the ball just before hitting the ground will be 36.4m/s