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

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

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

double the distance = divide strength by 4