Question

A rock is thrown horizontally from the top of the cliff 88m high, with a horizontal velocity of 25m/s. For what interval time is the rock in the air?

Answer 1

To find the time interval during which the rock is in the air, we can use the equation of motion:

$h=\frac{1}{2}g{t}^2$

where $h$ is the vertical displacement (88 m in this case), $g$ is the acceleration due to gravity (approximately 9.8 m/s${}^2$), and $t$ is the time.

Since the rock is thrown horizontally, the initial vertical velocity is zero. Thus, we can solve for $t$ in the above equation.

We can rewrite the equation as: $t=\sqrt{\frac{2h}{g}}$

Substituting the given values, we have: $t=\sqrt{\frac{2\times 88}{9.8}}$

Calculating this expression: $t\approx$