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

Answer:

Given :-

• A ball is dropped from a bridge of 122.5 m above a river.
• After the ball has been falling for two seconds, a second ball is thrown straight down after it.

To Find :-

• What is the initial velocity of second ball so that both hit the water at the same time.

Solution :-

First, we have to find the time taken by the first object to reach the ground :

Given :

• Height = 122.5 m
• Acceleration due to gravity = 9.8 m/s²
• Initial Velocity = 0 m/s

As we know that :

$\clubsuit$ Second Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{h =\: ut + \dfrac{1}{2}gt^2}}\: \: \: \bigstar$

where,

• h = Height
• u = Initial Velocity
• t = Time Taken
• g = Acceleration due to gravity

According to the question by using the formula we get,

$\implies \bf h =\: ut + \dfrac{1}{2} gt^2$

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

$\implies \sf 122.5 =\: 0 + \dfrac{1}{2} \times 9.8 \times t^2$

$\implies \sf 122.5 =\: 0 + \dfrac{9.8}{2} \times t^2$

$\implies \sf 122.5 - 0 =\: \dfrac{9.8}{2} \times t^2$

$\implies \sf 122.5 =\: \dfrac{9.8}{2} \times t^2$

$\implies \sf 122.5 \times \dfrac{2}{9.8} =\: t^2$

$\implies \sf \dfrac{245}{9.8} =\: t^2$

$\implies \sf 25 =\: t^2$

$\implies \sf \sqrt{25} =\: t$

$\implies \sf 5 =\: t$

$\implies \sf\bold{\underline{t =\: 5\: seconds}}$

Hence, the time taken by the first object to reach the ground is 5 seconds .

Now, we have to find the time taken by the second ball to reach the ground :

$\leadsto \sf Time\: Taken =\: 5 - 2$

$\leadsto \bf Time\: Taken =\: 3\: seconds$

Now, again we have to find the initial velocity of second ball :

Given :

• Height = 122.5 m
• Acceleration due to gravity = 9.8 m/s²
• Time Taken = 3 seconds

According to the question by using the formula we get,

$\implies \bf h =\: ut + \dfrac{1}{2} gt^2$

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

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

$\implies \sf 122.5 =\: 3u + \dfrac{1}{2} \times 88.2$

$\implies \sf 122.5 =\: 3u + \dfrac{88.2}{2}$

$\implies \sf 122.5 =\: 3u + 44.1$

$\implies \sf 122.5 - 44.1 =\: 3u$

$\implies \sf 78.4 =\: 3u$

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

$\implies \sf 26.13 =\: u$

$\implies \sf\bold{\underline{u =\: 26.13\: m/s}}$

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