Question

A STONE IS THROWN VERTICALLY UPWARD WITH THE VELOCITY 25/s HOW LONG DOES IT TAKES TO REACH THE MAXIMUM HEIGHT?

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Answer 1

time taken to reach maximum height is 2.548 sec and Maximum height is 31.855 m

Step-by-step explanation:

Initial velocity U= 25 m/s

at maximum height velocity, V= 0 m/s

Let the time taken to reach maximum height be t and maximum height be H.

V= U-gt

so 25-9.81.t=0

therefore t= 25/9.81 = 2.548 sec

Now,

H= Ut-1/2(gt^2)

=25x2.548-0.5x9.81x(2.548^2)

=31.855 m

Thank you