Question

A stone is projected vertically upwards with a velocity of 20 metre per sec. Two seconds later a second stone is similarly projected with the same velocity when the two stones meet the second one is rising at a velocity of 10 metre per sec. Neglecting air resistance, calculate the :
¡ Length of time the second stone is in motion before they meet.

¡¡ Velocity of the first stone when they meet ( take g= 10 metre per sec²)

Answer 1

I . v= u + at

0= 20 + 10 t

-10t = 20

t= 20/10

t= 2

now , we find distance

s= ut + half at*t

s= 20*2+ half 10*2*2

s=40+20s=60

s= 60m

2. v= u+ at

v= 10 + 10*2

v= 40