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A ball is thrown straight up from the edge of the roof of a building of height 60.0m. A second ball is dropped from the roof 2s later. What must be the initial speed of the first ball if both are to hit the ground at the same time?

Respuesta :

Answer:

u = 15.99 m/s

Explanation:

Given that

h= 60 m

For second ball

Lets take after t time after second ball hit the ground

[tex]h=ut+\dfrac{1}{2}gt^2[/tex]

Here u = 0 m/s

[tex]-60=-\dfrac{1}{2}\times 9.81\times t^2[/tex]

t=3.49 s

So the time taken by first ball will be 3.49 + 2 = 5.49 s

For first ball

[tex]-h=ut-\dfrac{1}{2}gt^2[/tex]

[tex]-60=u\times 5.49-\dfrac{1}{2}\times 9.81\times 5.49^2[/tex]

u = 15.99 m/s

So the initial velocity of first ball will be 15.99 m/s

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