Question

A particle when thrown, moves such that it passes from same height at 2 and 10 second, the height is (a) g (b) 2g (c) 5g (d) 10g

Answer 1

Answer:

H = V t - 1/2 g t^2      height after t sec

V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2       same height at t1 and t2

V (t2 - t1) = 1/2 g * (t2^2 - t1^2)

V = 1/2 g * (t2 + t1)      initial speed

V = 1/2 g * (10 + 2) = 6 g

H = V t - 1/2 g t^2 = 6 g * 2 - 1/2 g * 4) = 12 g - 2 g = 10 g

or

H = 6 g * 10 - 1/2 g * 100 = 10 g