First slide
Vertical projection from ground
Question

A body is projected vertically upward with a velocity of 50 m/s . Then the ratio of distances  travelled in the first second of upward motion to first second of downward motion is (Take g =10 m/s2).

Moderate
Solution

h1 = Distance travelled in the first second of upward motion = 50x1 - 1/2 x 10 x 12 = 45 m

ta = Time of ascent = u/g = 50/10 sec = 5 sec

So at the end of 5 sec , The body is at the highest position and from that position it starts falling freely

.: Distance travelled in the first second of downward motion , h2 = 1/2 x 10 x 12 m = 5m

.:h1h2 = 45/5 = 9

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