Question

A ball is projected vertically upwards. If it covers equal distances in 4th and 7th second, then velocity of projection is :-

Answer 1

Answer:

S3 = V t - 1/2 g t^2     where V is projection speed  and t = 3 sec

S3 = 3 V - 9 g / 2     height after 3 sec

S4 = 4 V - 8 g         height after 4 sec

S4 - S3 = V - (8 - 9/2) g = V - 7 g / 2

S6 = 6 V - 1/2 g * 36 = 6 V - 18 g     height after 6 sec

S7 = 7 V - 49 g / 2       height after 7 sec

S4 - S3 = S6 - S7      equal distances

V - 7 g / 2  = -V + g (-18 + 49 /2) = -V + 13 g / 2

2 V = 20 g / 2  

V = 5 g =  160 m /s

After 3 sec V = 160 - 96 = 64 m/s

after 4 sec V = 160 - 128 = 32 m/s

after 6 sec V = 160 - 192 m/s = = -32 m/s

after 7 sec V = 160 - 224 m/s = -64 m/s

Using 2 g S = V2^2 - V1^2 the distances are equal