Question

A stone is dropped freely in a river from a bridge. It takes 5s to touch the water surface in the river. Calculate : (i) the height of the bridge from the water level, (ii) the distance covered by the stone in the first 2 sec (g = 9.8 m s-2)

Answer 1

Answer:

• 122.5 m.
• 19.6 m.

Explanation:

Given: u = 0 m/s, a = g = 9.8 m/s² , t = 5 s.

To find: Height & distance covered.

Solution:

(i) From equation of motion, h = ut + 1/2gt²

⇛ h = 0 × 5 + 1/2 × 9.8 × 5²

⇛ h = 9.8 × 25/2

⇛ h = 122.5 m

∴ Height of the bridge is 122.5 m.

(ii) Distance covered by the stone in t = 2s

∅ S = ut + 1/2gt²

⇛ S = 0 + 1/2 × 9.8 × 2²

⇛ S = 19.6 m.

∴ The distance covered by the stone is the first 2 s is 19.6 m.

Answer 2

QUESTION:-

A stone is dropped freely in a river from a bridge. It takes 5s to touch the water surface in the river. Calculate : (i) the height of the bridge from the water level, (ii) the distance covered by the stone in the first 2 sec (g = 9.8 m/s²)

EXPLANATION:-

i)Height of bridge

To calculate the height of the bridge we will apply 2nd equation of motion

So the second equation of motion is (under free fall)

$\boxed{\bf\;H=ut+\frac{1}{2}at^2}}$

Where

H=height

u=initial velocity

t=time taken

a=acceleration due to gravity

So here,

u=0 m/s   (as ball falls from rest)

t=5 s

a=9.8 m/s²

So let's put the given data in the formula to find the height

h = 0 × 5 + 1/2 × 9.8 × 5²

 h = 9.8 × 25/2

⇛ h = 122.5 m

So height of bridge is 122.5 m

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(ii) the distance covered by the stone in the first 2 sec

To find the distance travelled in first 2 second we will again apply 2nd equS h= ut + 1/2gt²

 S = 0 + 1/2 × 9.8 × 2²

⇛ S = 19.6 m

So distanace travelled in first 2 second is 19.6 m

______________________________

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