Question

a ball dropped from height h reaches the ground in t secs. after what timr the ball was passing through a point at a height h/2​

Answer 1

Here we have,

The height from which the ball is dropped = h

The time taken by the ball to reach on ground = T

Initial velocity of the ball is zero because the ball is dropped not thrown.

and, acceleration due to gravity (g) = 9.8m/s²

Now  by using the 2nd equation of motion that is h = 1/2gT²+ uT, we get

                      h = ½ × 9.8 ×T² + 0 × T

                      h = 4.9T²  

                      T² = h/4.9        ……..eq(i)

Now let the distance(height)  covered by the ball in time 3T/4 be 'H'

Using the 2nd equation of motion that is H = 1/2gT²+ uT, we get                

                       H = ½ × 9.8 × (3T/4)² + 0 × (3T/4)

                       H =  4.9 × 9T²/16

   Putting the value of T² from eq(i), we get

                      H = 4.9 ×9/16 × h/4.9

                     H =  9h/16

Thus, the height of ball at time 3T/4 is (h - H)

                       =  h - 9h/16

                       = 16h - 9h /16

                       =  7h/16

Hence, 7h/16 is the our answer of this question.

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