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:

Given that:-

S = – 60 m

u = 5 m/s

a = 10 m/s (assuming)

where s = height

u = initial velocity

and a = acceleration due to gravity

To find:-

Height of balloon when stone hits the ground

Solution:-

Now using 2nd equation of motion,

s = ut + 1/2 at$^{2}$

Substitute the known values

$\Longrightarrow \: - 60 = 5(t) + \frac{1}{2} \times ( - 10) \: t {}^{2} \\ \\ \Longrightarrow \: - 60 = 5t - 5t {}^{2} \\ \\ \Longrightarrow 5t {}^{2} - 5t - 60 = 0 \\ \\ \Longrightarrow \: t {}^{2} - t - 12 = 0 \\ \\ \Longrightarrow \: t {}^{2} - 4t + 3t - 12 = 0 \\ \\ \Longrightarrow \: (t - 4)(t - 3) = 0 \\ \\ \Longrightarrow \: t = 4$

t = 4

So, its height when stone hits the ground =

60 + 4 × 5 = 60 + 20

= 60 m.

Answer 2

Answer:

60M Will be the correct answer hope it helps you