Question

a partical is thrown vertically upwards moves such that it passes form the same heighyt at t=2 and t=10 the height is?

Answer 1

Answer:

h = V0 t - 1/2 g t^2          where h is the height after time t

V0 t1 - 1/2 g t1^2 = V0 t2 - 1/2 g t2^2     since height is the same

2 V0 (t2 - t1) = g (t2^2 - t1^2)

2 V0 = g (t2 + t1)     since  a^2 - b^2 = (a + b) * (a - b)

V0 = 1/2 * 9.8 * 12 = 58.8 m/s    initial upward speed

h = 58.8 * 2 - 1/2 * 9.8 * 2^2 = 98 m

h = 58.8 * 10 - 1/2 * 9/8 * 10^2 = 98 m