Question

With what speed in Km/h must an object be thrown to reach a height of 91.5 m? Assume negligible air resistance.

Answer 1

The speed is `=152.5kmh^-1`

Explanation:

Resolving in the vertical direction `uarr^+`

Apply the equation

`v^2=u^2+2as`

The acceleration due to gravity is `a=g=-9.8ms^-2`

The height is `s=h=91.5m`

The final velocity is `v=0ms^-1`

Therefore,

`0=u^2-2xx9.8xx91.5`

`u^2=2*9.8*91.5=1793.4`

`u=sqrt(1793.4)=42.35ms^-1`

`=42.35*3600/1000`

`=152.5kmh^-1`

Answer 2

152.53 km/h

Explanation:

Recalling one of the equations of motion:

`v^2 = u^2 + 2as`

Where `u` is the initial velocity before and `v` is the final velocity after a free fall acceleration.

Since we are talking about a deceleration, we can use this same equation to calculate the throwing velocity `v` while assuming that `u,` the velocity when the ball is at its highest point, is clearly zero.

The main idea here is that the velocity you will need to throw the ball at and the velocity at which this ball will hit the ground again are the same, so this equation applies in both cases.

Substituting zero for `u` and 9.81 (metres per second squared) for `a,`

`v^2 = 2* 9.81 * 91.5`
`v^2= 1795.23`
`v = 42.37`

Therefore, `v` is 42.37 m/s and 152.53 km/h.