# a ball is dropped from a bridge of 122.5 metre above a river. after the ball has been falling for two seconds, a second ball is thrown straight down after it. initial velocity of second ball so that both hit the water at the same time is

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Question

A ball is dropped from a bridge

122.5 m

high. After the first ball has fallen for

2

second, a second ball is thrown straight down after it, what must be the initial velocity of the second ball, so that both the balls hit the surface of water at the same time?

Open in App Solution

The correct option is **A**

26.1 m/s

Time taken by the first object to reach the ground

= t , so 122.5 = u t + 1 2 g t 2 122.5 = 1 2 × 10 × t 2 ⇒ t = 5 sec(approx)

Time to be taken by the second ball to reach the ground

= ( 5 − 2 ) sec = 3 sec . If u

be its initial velocity then,

122.5 = u × 3 + 1 2 g t 2 = 3 u + 1 2 × 10 × 9 ⇒ 3 u = 122.5 − 45 = 77.5 ∴ u = 26 (approx).

Motion Under Gravity

Standard XII Physics

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## A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of second ball so that both hit the water at the same time is

Click here👆to get an answer to your question ✍️ A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of second ball so that both hit the water at the same time is

Question

## A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of second ball so that both hit the water at the same time is

**A**

## 49m/s

**B**

## 55.5m/s

**C**

## 26.1m/s

**D**

## 9.8m/s

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

For the first ball: Let it takes′ t ′ sec to river. ⇒122.5= 2 1 ×g×t 2

------------------------(1)$

for second ball: time taken =t−2sec.

122.5=u(t−2)+ 2 1 g(t−2) 2 --------------(2) ⇒ 2 gt 2 =ut−2u+ 2 gt 2 +2g−2gt ⇒(2g−u)t=2g−2u ⇒t= 2g−u 2g−2u from(1) t 2 = g 2×122.5 t= 98 2×122.5 = 7 35 =5sec ⇒10g−5u=2g−2u ⇒8g=3u u= 3 8×9.8 =26.1m/s Hence,

option (C) is correct answer.

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## The ball is dropped from a bridge 122.5 m above a river, After the ball has been falling for 2 s, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time ?

Let the first ball hit the qater in t seconds Using , S= ut + 1/2 at^2 For first ball, 122.5 =0 + 1/2 xx 9.8 xx t^2 = 4.9 T^2 ltbRgt or t= sqrt (122.5)/(4.9) = 5 s ltbRgt For second ball, 122.5 = u (5 -2) + 1/2 xx 9.8 xx (5 - 2)^2 = 2 u + 44 .1 3 u = 122.5 + 44 .1 3 u = 122.5 - 44.1 = 78.4 or u=26.1 ms^(-1).

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The ball is dropped from a bri...

The ball is dropped from a bridge

122.5m 122.5m

above a river, After the ball has been falling for 2 s, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time ?

Updated On: 27-06-2022

00 : 30

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Text Solution Open Answer in App A (a) 40m/s 40m/s B (b) 55.5m/s 55.5m/s C (c ) 26.1m/s 26.1m/s D (d) 9.6m/s 9.6m/s Answer

The correct Answer is C

Solution

Let the first ball hit the qater in

t t seconds Using , S=ut+ 1 2 a t 2 S=ut+12at2 For first ball, 122.5=0+ 1 2 ×9.8× t 2 =4.9 T 2

122.5=0+12×9.8×t2=4.9T2

ltbRgt or t= 122.5 − − − − √ 4.9 =5s t=122.54.9=5s

ltbRgt For second ball,

122.5=u(5−2)+ 1 2 ×9.8× (5−2) 2

122.5=u(5-2)+12×9.8×(5-2)2

=2u+44.1 =2u+44.1 3u=122.5+44.1 3u=122.5+44.1

3u=122.5−44.1=78.4oru=26.1m

s −1

3u=122.5-44.1=78.4oru=26.1ms-1

.

Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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Guys, does anyone know the answer?