Question

During a thunderstorm, a tornado lifts a car to a height of 125 m abovethe ground. Increasing in strength, the tornado flings the car horizon-tally with a speed of 90.0 m/s. How long does thecar take to reach the ground? How far horizontal-does the car travel before hitting the ground?

Answer 1

The time taken by car to reach the ground is,

`t=\sqrt[]{(2h)/(g)}`

Substitute the known values,

`\begin{gathered} t=\sqrt[]{\frac{2(125\text{ m)}}{(9.8m/s^2)}} \\ \approx5.05\text{ s} \end{gathered}`

Therefore, the time taken by car to reach the ground is 5.05 s.

The distance covered by car is given as,

`d=vt`

Substituting values,

`\begin{gathered} d=(90.0\text{ m/s)(5.05 s)} \\ =454.5\text{ m} \end{gathered}`

Thus, the horizontal distance travelled by car is 454.5 m.