A ball is projected vertically up with an initial speed of `20m//s` on a planet where acceleration due to gravity is `10m//s^(2)` (a) How long does it takes to reach the highest point? (b) How high does it rise above the point of projection? (c) How long wil it take for the ball to reach a point `10m` above the point of projection?

A ball is projected vertically up with an initial speed of `20m//s` on

1.	A ball is projected vertically up with an initial speed of `20m//s` on a planet where acceleration due to gravity is `10m//s^(2)` (a) How long does it takes to reach the highest point? (b) How high does it rise above the point of projection? (c) How long wil it take for the ball to reach a point `10m` above the point of projection?
Answer» As here motion is vertically upwards, `a =- g` and `v = 0` (a) From `1st` equaiton of motion, i.e., `v = u +at`, `0 = 20-10t rArr t = 2sec.` (b) Using `v^(2) =u^(2) +2as` `0 = (20)^(2) -2x 10x h rArr h = 20m`. (c ) Using `s = ut +(1)/(2)at^(2), rArr` `10 =20t (-(1)/(2)) x10xt^(2) rArr t^(2)-4t+2 =0` `rArr t = 2+- sqrt(2), rArr t = 0.59 sec.` or `3.41sec.` i.e., there are two times, at which the ball passes thorugh `h = 10m`, once while going up and then coming down.

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