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 :-
Share
Verified answer
Answer:
Given :-
To Find :-
Solution :-
First, we have to find the time taken by the first object to reach the ground :
Given :
As we know that :
[tex]\clubsuit[/tex] Second Equation Of Motion Formula :
[tex]\bigstar \: \: \sf\boxed{\bold{h =\: ut + \dfrac{1}{2}gt^2}}\: \: \: \bigstar\\[/tex]
where,
According to the question by using the formula we get,
[tex]\implies \bf h =\: ut + \dfrac{1}{2} gt^2\\[/tex]
[tex]\implies \sf 122.5 =\: (0)t + \dfrac{1}{2} \times (9.8) \times t^2\\[/tex]
[tex]\implies \sf 122.5 =\: 0 + \dfrac{1}{2} \times 9.8 \times t^2\\[/tex]
[tex]\implies \sf 122.5 =\: 0 + \dfrac{9.8}{2} \times t^2\\[/tex]
[tex]\implies \sf 122.5 - 0 =\: \dfrac{9.8}{2} \times t^2\\[/tex]
[tex]\implies \sf 122.5 =\: \dfrac{9.8}{2} \times t^2\\[/tex]
[tex]\implies \sf 122.5 \times \dfrac{2}{9.8} =\: t^2\\[/tex]
[tex]\implies \sf \dfrac{245}{9.8} =\: t^2\\[/tex]
[tex]\implies \sf 25 =\: t^2\\[/tex]
[tex]\implies \sf \sqrt{25} =\: t[/tex]
[tex]\implies \sf 5 =\: t[/tex]
[tex]\implies \sf\bold{\underline{t =\: 5\: seconds}}\\[/tex]
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 :
[tex]\leadsto \sf Time\: Taken =\: 5 - 2\\[/tex]
[tex]\leadsto \bf Time\: Taken =\: 3\: seconds\\[/tex]
Now, again we have to find the initial velocity of second ball :
Given :
According to the question by using the formula we get,
[tex]\implies \bf h =\: ut + \dfrac{1}{2} gt^2\\[/tex]
[tex]\implies \sf 122.5 =\: u(3) + \dfrac{1}{2} \times (9.8)(3)^2\\[/tex]
[tex]\implies \sf 122.5 =\: 3u + \dfrac{1}{2} \times 9.8 \times (3 \times 3)\\[/tex]
[tex]\implies \sf 122.5 =\: 3u + \dfrac{1}{2} \times 88.2\\[/tex]
[tex]\implies \sf 122.5 =\: 3u + \dfrac{88.2}{2}\\[/tex]
[tex]\implies \sf 122.5 =\: 3u + 44.1[/tex]
[tex]\implies \sf 122.5 - 44.1 =\: 3u[/tex]
[tex]\implies \sf 78.4 =\: 3u[/tex]
[tex]\implies \sf \dfrac{78.4}{3} =\: u[/tex]
[tex]\implies \sf 26.13 =\: u[/tex]
[tex]\implies \sf\bold{\underline{u =\: 26.13\: m/s}}\\[/tex]
[tex]\therefore[/tex] The initial velocity of second ball so that both hit the water at the same time is 26.13 m/s .