Question

A ball projected from ground vertically upward is at same height at times t1 nd t2. The speed of projection of ball is??

Answer 1

Given:

Let h be the height attained by both balls.

Let u be the projection of ball

time of first ball=t1

time of second ball=t2

For first ball :

s=ut+1/gt²
h=ut1-1/2 gt1² ---------1

For second ball:

h=ut2-1/2 gt2²----------2
As same height is attained:
equating equation 1 and 2

ut1-1/2gt1²=ut2-1/2 gt2²

ut1-ut2=1/2 gt1²-1/2gt2²

u(t1-t2)=1/2g(t1²-t2²)

u=g[t1²-t2²]/2(t1-t2)

u=g[t1-t1][t1+t2]/2(t1-t2)

u=g[t1+t2]/2

∴The speed of projection of ball is g[t1+t2]/2

Answer 2

Answer:g(t1+t2)/2

Explanation:

Since particle is at same height at time t1&t2

Therefore,

(t1 + t2)= total time of flight

(t1 + t2) =2T

T=(t1+t2)/2

AlsoT=u/g from (v=u+at and v being 0 at maximum height and g is negative)

Therefore,

u/g=(t1+t2)/2

u=g(t1+t2)/2

HOPE IT WILL HELP YOU