A juggler performs in a room whose ceiling is 3.0 meters above the level of his hands he throws a ball upward so that it just reaches the ceiling what is the initial velocity of the ball?

Answer 1

First summarise what we know as equations and values.

Call the initial velocity of the ball vi. This is the value we need to calculate.

We know:

The final velocity of the ball vf = 0m.s-1. It has completely stopped at the top of its travel. (We are not interested in the trip down.)

The deceleration of the ball is caused by gravity (discounting air friction) which is 9.81m.s-2. So the acceleration a = -9.81m.s-2. It is negative because it is acting in the opposite direction to the initial movement of the ball.

The distance travelled by the ball before it comes to a stop at the top is d = 3m.

So we have: vf, a and d, and we need an equation with these and vi that we can arrange to solve for vi.

We know from basic kinematic equations that:

# vf = vi + a.t # d = vi.t + 1/2.a.t2

but these require the time value t which we do not know.

However, because we have two equations, we can solve one for t to eliminate it:

rearranging Equation 1 to solve for t gives us:

# t = (vf - vi)/ a and we can substitute this into Equation 2 and, after some rearranging to solve for vi, which is the value we want, get:

vi = the square root of (vf2 - 2.a.d)

So putting in the values gives us:

vi = the square root of (02 - 2 . -9.81 . 3)

= the square root of ( - 2 . -9.81 . 3)

= the square root of ( - 6 . -9.81 )

= the square root of ( 58.86)

= 7.67

= 7.7m.s-1