Question

A cannonball of mass 1kg is shot vertically upward from the top of a building with an unknown velocity v_0(m/sec).v 0 ​ (m/sec). After 2 seconds, the ball reaches its maximum height 30 meters from the ground. Neglecting the air resistance and taking the gravitational acceleration gg to be 10(m/sec^2)10(m/sec 2 ), find the height of the building.

Answer 1

Taking the upward direction to be positive, the cannonball's height
`y(t)` in the air at time
`t` is given by

`y(t)=y_0+v_0 t-\frac g2t^2`

where
`g` is the magnitude of the acceleration due to gravity, 10 m/s^2, and
`y_0` is the height of the building from which the ball is being thrown.

At the moment the cannonball reaches its maximum height of 30 m, its velocity at that time is 0, so that

`0^2-{v_0}^2=-2g(30\,\mathrm m-y_0)\implies v_0=\sqrt{\left(20(\rm m)/(\mathrm s^2)\right)(30\,\mathrm m-y_0)}`

Substitute this into the height equation above, and let
`t=2\,\mathrm s`, for which we have
`y(2\,\mathrm s)=30\,\mathrm m`:

`30\,\mathrm m=y_0+\sqrt{\left(20(\rm m)/(\mathrm s^2)\right)(30\,\mathrm m-y_0)}(2\,\mathrm s)-\left(5(\rm m)/(\mathrm s^2)\right)(2\,\mathrm s)^2`

Solve for
`y_0`: (units omitted for brevity; we know that
`y_0` should be given in m)

`30=y_0+4√(150-5y_0)-20`

`50-y_0=4√(150-5y_0)`

`(50-y_0)^2=\left(4√(150-5y_0)\right)^2`

`2500-100y_0+{y_0}^2=16(150-5y_0)`

`{y_0}^2-20y_0+100=0`

`(y_0-10)^2=0`

`\implies\boxed{y_0=10\,\mathrm m}`