A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the velocity just before it hits the water?

 v = 42.92 m/s

Explanation:

Given,

initial speed of the ball, v = 11 m/s

time taken to hit the ground = 5.5 m/s

velocity of the ball just before it hit the ground, v = ?

time taken by the ball to reach the maximum height

using equation of motion

 v = u + at

final velocity = 0 m/s  

 0 = 11 - 9.8 t

  t = 1.12 s.

time taken by the ball to reach the water from the maximum height

 t' - 5.5 -1.12 = 4.38 s

using equation of motion for the calculation of speed just before it hit the water.

 v  = u + a t

 v = 0 + 9.8 x 4.38

 v = 42.92 m/s

Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

answer: a. same unit

explanation: