A stone is thrown vertically upwards with a velocity of 19.6 m/s.After 2 sec,another stone is thrown upwards with a velocity of 9.8 m/s.When and where will these stones collide?
Answers
Answer:
They will meet at the point of projection after 4s of throwing the first stone.
Explanation:
Hope it helps!
Answer:
4 seconds
Explanation:
(Let's take upwards as positive direction. so acceleration due to gravity is
-g = -9.8 m/s^2)
After two seconds, velocity of the first ball is given by first equation of motion, v=u+at where v is final velocity, u is initial velocity, a is acceleration, t is time
v1=19.6 - 9.8*2 -------------------- (substituting values)
= 0
Height above the ground is given by second equation of motion
S = ut + 1/2at^2 where S is the displacement
= 19.6*2 - 9.8*2*2*1/2 = 19.6*2 - 19.6 = 19.6 m
At that time velocity of ball 2 is 9.8 m/s
Now relative acceleration of ball 1 wrt (with respect to) ball 2 is given by : acceleration of first ball - acceleration of second ball
= -g - (-g) = 0
Relative velocity of ball 1 wrt ball 2 is given by :
velocity of first ball - velocity of second ball
= 0 - 9.8 = -9.8 (minus sign is because the relative velocity of ball 1 will be downwards wrt ball 2)
Since relative acceleration is zero, this will be the relative velocity till there collision.
Now distance between them is 19.6 (as ball 2 is on the ground, ball 1 is at a height of 19.6 m)
So time is simply distance/speed = 19.6/9.8 = 2 seconds
Therefore total time is 2+2 = 4 seconds