A boy standing at the top of a tower of $30 m$ height drops a stone. Assuming $g = 10 {ms}^{- 2}$ the magnitude of velocity with which it hits the ground is

20 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

24.49 m / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

40 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.26 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $24.49 m / s$
Given, $g = 10 {ms}^{- 2}$ and $h = 30 m$
Taking downward as positive,
Using third equation of motion, we have
$v^{2} - u^{2} = 2 g h$
$\Rightarrow v = \sqrt{2 g h} (∵ u = 0)$
$= \sqrt{2 \times 10 \times 30}$
$= \sqrt{600} = 24.49 {ms}^{- 1}$

Suggest Corrections

Similar questions

Q. A boy standing at the top of a tower of

20 m

height drops a stone. Assuming

g = 10 {ms}^{- 2}

the magnitude of velocity with which it hits the ground is

A boy standing at the top of a tower, which is 20m high, drops a stone. Assume g = 10 $\frac{m}{s^{2}}$ , find the velocity with which the stone hits the ground.

A boy standing at top of a tower of 20 m height drops a stone. Assuming g=10m/S2. The velocity with which it hits the ground is???

A fruit is dropped from a height of 20 m. Find the velocity with which it would hit the ground. Take g = $10 m s^{- 2}$

A fruit is dropped from a height of 20 m. Find the velocity with which it would hit the ground.
Assume acceleration due to gravity $g = 10 m / s^{2}$ .

Join BYJU'S Learning Program

A boy standing at the top of a tower of 30 m height drops a stone. Assuming g=10 ms−2 the magnitude of velocity with which it hits the ground is

The correct option is B 24.49 m/sGiven, g=10 ms−2 and h=30 m Taking downward as positive, Using third equation of motion, we have v2−u2=2gh ⇒v=√2gh (∵u=0) =√2×10×30 =√600=24.49 ms−1

A boy standing at the top of a tower of $30 m$ height drops a stone. Assuming $g = 10 {ms}^{- 2}$ the magnitude of velocity with which it hits the ground is

The correct option is B $24.49 m / s$
Given, $g = 10 {ms}^{- 2}$ and $h = 30 m$
Taking downward as positive,
Using third equation of motion, we have
$v^{2} - u^{2} = 2 g h$
$\Rightarrow v = \sqrt{2 g h} (∵ u = 0)$
$= \sqrt{2 \times 10 \times 30}$
$= \sqrt{600} = 24.49 {ms}^{- 1}$