Answer:
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
