Question

A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 60 m high when the stone was dropped, find its height when the stone hits the ground.​

Answer 1

Answer:

BY APPLYING FORMULA:

By taking g=10m/s²

$s = ut + \frac{1}{2}a {t}^{2} \\ = > - 60 = 5(t) + \frac{1}{2}(-10) {t}^{2} \\ = > - 60 = 5t - 5 {t}^{2} \\ = > 5 {t}^{2} - 5t - 60 = 0 \\ = > {t}^{2} - t - 12 = 0 \\ = > {t}^{2} - 4t + 3t - 12 = 0 \\ = > t(t - 4)+3(t-4)=0 \\ = > (t - 4)(t + 3) = 0$

Here we'll have two cases:

1. When (t-4)=0, then t=4
2. When (t+3)=0, then t=-3

But in case 2 't' is negative, which is not possible.

So case 1 is correct.

Therefore:

t=4

Now:

Height of the balloon from the ground at this instant=60+4×5=80m

So there's your answer buddy ✌️

Refer to the attachment for better understanding ❣️