Question

A particle dropped from a balloon rising at 20ms^-1 at a time when the balloon is 80m above the ground. If g=10ms^-2, then the particle reaches the ground approximately in
1)5.6s
2)6.5s
3)4.5s
4)4s​

Answer 1

Answer:

Explanation:

(All the signs are taken by assuming +y axis as + and -y axis as -)

The initial velocity of the particle is 20 m/s along +y axis

we know that $s = ut + \frac{at^{2} }{2}$

substituting u = 20 m/s, a = -g = -10 m/s² , s = -80(as the particle goes down ultimately) , we get,

$-80 = 20t - 5t^{2}$

this is a quadratic in t

$5t^{2} - 20t - 80 = 0$

applying the quadratic formula and disregarding the negative root as time cannot be negative we get,

t = 2 + $\sqrt{20}$ ≈ 6.5 s