Question

You hit a golfball vertically upward with your pitching wedge. The position function of the ball is s(t) = -3t^2+30t where the origin is at ground level and the positive direction is vertically upward. (Distance is in metres). Help!?

A) Find the maximum height reached by the ball
B) Find the velocity of the ball as it reaches the ground
C) Find the acceleration of the ball.

Answer 1

A) Maximum height is `75 \ m`
B) Velocity at ground level is `-30 \ ms^(-1)`
C) Acceleration is `= -6 \ ms^(-2)`

Explanation:

The vertical displacement (relative to ground level) of the ball is given by:

`s(t) = -3t^2+30t`

The velocity of the ball is given by the rate of change of displacement wrt time, ie:

`v(t) = d/dt s(t)`

So if we differentiate `s(t)` wrt `t` we get:

`v(t) = -6t+30`

Just prior to the maximum trajectory, the ball's velocity will increase then it will momentarily stop before starting to drop again, ie we seek `v(t)=0` (or, alternatively, given the definition of displacement, we seek a turning point and set the derivative to zero):

`v(t) = 0 => -6t+30 = 0`
`:. 6t = 30`
`:. t = 5`

When `t=5`, the displacement is:

`s(5) = -3xx25+30xx5`
`" " = -75+150`
`" " = -75+150`
`" " = 75`

So the maximum height is `75 \ m`

At ground level we have no displacement, ie `s(t)=0`

`s(t) = 0 => -3t(t-10) = 0`
# :. t=0, 10 ' (units)

We expected `t=0` as the initial condition, so `t=10` corresponds to the ball reaching ground level after flight.

`t = 10 => v(10) = -60+30 = -30 \ ms^(-1)`

We also expect a negative velocity as the ball is falling not rising

The acceleration of the ball is given by the rate of change of velocity wrt time, ie:

`a(t) = d/dt v(t) = (d^2)/(dt^2) s(t)`

So if we differentiate `v(t)` wrt `t` we get:

`a(t) = -6 \ ms^(-2)`, which is constant acceleration throughout flight