Question

A particle is dropped from a balloon rising at 20 m s1 at a time when the balloon is 80 m above the ground. if g = 10 m s2, then the particle reaches the ground approximately in

Answer 1

Answer:

The particle reaches the ground approximately in 4 seconds

Explanation:

In the question, it is given that...

s=80m

a=g= $10$ $m/s^2$

Substituting in the equation $s=ut+\frac{1}{2} at^2$,

$80=0*t+\frac{1}{2}*10*t^2\\\\80=0+5t^2\\80=5t^2\\\frac{80}{5}=t^2\\16=t^2\\\sqrt16=\sqrt(t^2)$

$t=4$