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 :- ​

Answer 1

For the first ball: Let it takes ′t′ sec to river.

⇒122.5= 1/2×g×t²-----------------------(1)$

for second ball: time taken =t−2sec.

122.5=u(t−2)+1/2g(t−2)²--------------(2)

⇒gt²/2=ut−2u+gt²/2+2g−2gt

⇒(2g−u)t=2g−2u

⇒t=2g−2u/2g-u

from(1)

t²=2×122.5/g

t=√2×122.5/98=35/7=5sec

⇒10g−5u=2g−2u

⇒8g=3u

u= 8×9.8/3 =26.1m/s

Answer 2

Given

• A ball is dropped from a height of 122.5 meters.
• After the ball has been falling for two seconds, a second ball is thrown down.
• First ball and second ball falls down at the same time.

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To Find

• The initial velocity of the second ball.

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Solution

Height of bridge (s) = 122.5 seconds

Initial Velocity of First Ball (u) = 0 m/s

Acceleration (a)  = 9.8 m/s² (Since it's experiencing free fall)

To find the time taken for the first ball to reach the ground, apply the second equation of motion.

$\boxed{\sf Second \ Equation \ of\ Motion -: s = ut + \frac{1}{2}at^{2} }$

Substitute the values of variables in the equation.

$\implies s = ut + \dfrac{1}{2}at^{2}$

$\implies 122.5 = 0(t) + \dfrac{1}{2} (9.8)t^{2}$

$\implies 122.5 = 4.9t^{2}$

$\implies t^{2} = \dfrac{122.5}{4.9}$

$\implies t^{2} = 25$

$\implies \bf t = 5 \ seconds$

∴ Time taken for the first ball to reach the ground is 5 seconds.

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Time taken for the second ball to reach the ground = Time taken for the first ball to reach the ground - 2

⇒ 5 - 2

⇒ 3 seconds

∴ Time taken for the second ball to reach the ground is 3 seconds.

To find the initial velocity of the second ball, apply the second equation of motion.

$\boxed{\sf Second \ Equation \ of\ Motion -: s = ut + \frac{1}{2}at^{2} }$

Substitute the values of variables in the equation.

$\implies s = ut + \dfrac{1}{2}at^{2}$

$\implies 122.5 = u(3) + \dfrac{1}{2}(9.8)(3)^{2}$

$\implies 122.5 = 3u + 4.9(9)$

$\implies 122.5 = 3u + 44.1$

$\implies 3u =122.5- 44.1$

$\implies 3u =78.4$

$\implies u = \dfrac{78.4}{3}$

$\implies \bf u = 26.13 \ m/s$

∴ The initial velocity of the second ball so that both hit the water at the same time is 26.13 m/s.

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