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?

Options

A. 2 s, 20 m, 3.41 s
B. 5 s, 20 m, 3.41 s
C. 2 s, 10 m, .59 s
D. 2 s, 5 m, .59 s

Explanation

As here motion is vertically upwards a = – g and v = 0
(A) From 1st equation of motion
i.e., v = u + at
` rArr 0 = 20 - 10 t, i.e., t = 2s `
(B) From 3rd equation of motion
i.e., ` v^(2) = u^(2) + 2as `
` rArr 0 = (20)^(2) - 2 xx10 h, i.e., h = 20 m `
(C) From 2nd equation of motion,
i.e., `s = ut + (1/2) at^(2)`
`rArr 10 = 20 t - (1//2) xx 10 xx t^(2)`
` t^(2) - 4t + 2 = 0 , i.e. t = 2 pm sqrt(2)`
i.e. t = 0.59 s or 3.41 s
i.e. there are two such times, as the ball passes
twice through h = 10 m once when going up and
once when coming down.
Hence correct answer is (A)