Question

30. A ball is dropped from a bridge at a height of 180 m over a river. After 2 s, a second ball is thrown
straight downwards. What should be the initial velocity of the second ball so that both hit the water
simultaneously?
OA) 2.5m/s​

Answer 1

Time taken for the first ball to fall freely 122.5m will be, t=√2S/g=√2×122.5/9.8=5s. ltbRgt Thus, the second ball thrown after 2 seconds with velocity (u) should cover a distance 122.5m in 3 seconds.

Answer 2

Answer:

The initial velocity of the second ball is 25m/s.

Explanation:

Given the height from where the ball is dropped, h = 180 m

Let the time of dropping the first ball be $t sec$

As the ball is dropped, the initial velocity of first ball,  $u_1=0$  

Let the initial velocity of the second ball be $u_2 ~m/s$

Both the balls hit the water at the same time, thus the time of the second ball is $(t-2)sec$

Using the equation of motion under gravity​ for a freely falling body for the first ball, we get

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

For acceleration due to gravity, $g=10m/s^{2}$

$\implies 180=\dfrac{1}{2} \times 10\times t^{2}$

$\implies t^{2} =\dfrac{180}{5} =36$

$\implies t =\sqrt{36} =6s$

Using the same equation of motion for the second ball, we get  

$h=u_2(t-2)+\dfrac{1}{2} g(t-2)^{2}$

Substituting the respective values, we get

$180=u_2(6-2)+\dfrac{1}{2} \times 10(6-2)^{2}$

$\implies 180=u_2\times 4+5 \times 16$

$\implies u_2= \dfrac{180-80}{4}= \dfrac{100}{4}=25m/s$

Therefore, the initial velocity of the second ball is 25m/s.

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