Question

A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. The ball lands on the ground 2.9 s after it is launched. What is the height of the building (in m)?

Answer 1

Let
`y_0` be the height of the building and thus the initial height of the ball. The ball's altitude at time
`t` is given by

`y=y_0+\left(15(\rm m)/(\rm s)\right)\sin35^\circ\,t-\frac g2t^2`

where
`g=9.80(\rm m)/(\mathrm s^2)` is the acceleration due to gravity.

The ball reaches the ground when
`y=0` after
`t=2.9\,\mathrm s`. Solve for
`y_0`:

`0=y_0+\left(15(\rm m)/(\rm s)\right)\sin35^\circ(2.9\,\mathrm s)-\frac12\left(9.80(\rm m)/(\mathrm s^2)\right)(2.9\,\mathrm s)^2`

`\implies y_0\approx16.258\,\mathrm m`

so the building is about 16 m tall (keeping track of significant digits).