Question

A stone is thrown vertically upward.
The stone cross a
particular height twice
at time t1 and t2.Find the initial
speed of the throw ?​

Answer 1

Explanation:

Let u be the initial velocity and h be the maximum height attained by the stone.

v12=u2−2gh,

(10)2=u2−2×10×2h

100=u2−10h ....(i)

Again at height h,

v22=u2−2gh

(0)2=u2−2×10×h

u2=20h  ... (ii)

So, from Eqs. (i) and (ii) we have

100=10h

h=10m