Question

A particle is projected vertically upwards with velocity u from a point a when it returns to point of projection

Answer 1

Answer:

Explanation:

h=ut-(1/2)gt^2.

h=80m, g=10m/s^2. Then,

t^2-(2/g)ut+(2/g)h=0. Solving,

t={(2/g)u+_[(4/g^2)u^2–4(2/g)h]^(1/2)}/2

The difference between two roots is 6

[(1/25)u^2–64]^1/2=6. Squaring

(1/25)u^2–64=36

u^2=2500 or

u=50m/s^2. Then, total height,

H=u^2/2g=2500/20=125m