Physics, asked by tiwariasmit14, 1 year ago

A ball is thrown with a velocity of 20m/s from the top of the tower of height 25m .1=find how high will the ball rise.2=time taken by the ball in touching the ground​

Answers

Answered by prashantkaushik42
1

Answer:

Explanation:

v = 20 m/s             H   =  25 m

S = ball rise (height of ball covered from tower in upward direction)

                    S  = \frac{u^2}{2g}

                    S = \frac{20^2}{2 x 10}        g=10 m/ s^2

                     S = \frac{400}{20}

                     S = 20 m

Maximum height from ground = 25 + 20  == 45 m

T = total time to reach the ground

              T = \frac{u}{g} + \sqrt{\frac{u^2}{g^2} + \frac{2H}{g}}

here u/g is the time of ascent and the other under root value is the time of descent

           T = \frac{20}{10} + \sqrt{\frac{400}{100} +\frac{50}{10} }

          T = 5 sec

If you want these derivation then message me

Hope it helps!!!

Thanks

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