Question

A ball is thrown directly downward with an initial speed of 8.30 m/s, from a height of 29.2 m. After what time interval does it strike the ground?

Answer 1

The ball's height
`y` at time
`t` is given by

`y = 29.2\,\mathrm m - \left(8.30(\rm m)/(\rm s)\right) t - \frac12 gt^2`

where
`g=9.80(\rm m)/(\mathrm s^2)`.

Solve for
`t` when
`y=0`. Omitting the units, we have

`29.2 - 8.30t - \frac g2 t^2 = 0`

I'll solve by completing the square.

`29.2 - \frac g2 \left(t^2 + \frac{16.6}g t\right) = 0`

`29.2 - \frac g2 \left(t^2 + \frac{16.6}g t + (8.3^2)/(g^2)\right) = -\fracg2 * (8.3^2)/(g^2)`

`29.2 - \frac g2 \left(t + \frac{8.3}g\right)^2 = -(8.3^2)/(2g)`

`\frac g2 \left(t + \frac{8.3}g\right)^2 = 29.2 + (8.3^2)/(2g)`

`\left(t + \frac{8.3}g\right)^2 = \frac{58.4}g + (8.3^2)/(g^2)`

`t + \frac{8.3}g = \pm \sqrt{\frac{58.4}g + (8.3^2)/(g^2)}`

`t = -\frac{8.3}g \pm \sqrt{\frac{58.4}g + (8.3^2)/(g^2)}`

`\implies t \approx -3.43 \text{ or } t \approx 1.74`

Ignore the negative solution; the ball hits the ground about 1.74 s after being thrown.