Question

a hot air balloon is ascending vertically at a constant rate of 2.50m/s. When the balloon is 3 meters above the ground the pilot accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

Answer 1

The time elapses before the compass hits the ground is 1.0781s

Explanation:

Given:

Velocity, v = 2.5m/s

Height, h = 3m

Time elapse, t = ?

We know,

a = -9.8m/s²

$y = vt + \frac{1}{2} at^2\\\\t^2 + (\frac{2v}{a} )t - (\frac{2y}{a} ) = 0$

It is a quadratic equation and the solution would be:

$t = \frac{-(\frac{2v}{a}) +- \sqrt{(\frac{2v}{a} )^2 + 4 (1)(\frac{2y}{a}) } }{2}$

$t = - (\frac{v}{a}) +- \sqrt{(\frac{v}{a})^2 + (\frac{2y}{a} ) }$

The balloon goes from 3m to 0. So, y = -3m

On substituting the values in the equation we get:

$t = - (\frac{2.5}{-9.8} ) +- \sqrt{(\frac{2.5}{-9.8})^2 + (\frac{2(-3)}{-9.8} ) }$

t = 0.2551 ± 0.823

Since the time cannot be negative the value of t becomes 1.0781s

Therefore, time elapses before the compass hits the ground is 1.0781s