Mechanics questions?

Question

1) A ball is let to fall from a height of 40m. simultaneously and vertically bellow it, a second ball is thrown vertically upwards with a velocity of 20m/s at what distance from the ground do they collide?

2) Astone is thrown vertically upwards with a speed of 20m/s .one second later , a second stone is thrown vertically upwards with a speed of 30m/s .at what height above the ground do they collide?

Answer 1

venkat has the first part right but common sense tells us they must collide.

your two equations for part 20 should be:

for the first stone

s1=20t-0.5gt^2

for the second stone

s2=30(t-1)-0.5g(t-1)^2

equate s1=s2

eventually you will get t=1.763 seconds

plug back into either displacement equation and you get

s1 = s2 = 20.03 metres

I'm not sure what your last comment really means but they collide when their displacements are equal.

equating the displacement of both stones using s=ut-.5gts allows you to do this.  Bearing in mind that the second stone has time of (t-1)

Answer 2

Let the Point on ground vertically below the first ball be the origin.

then, the y-coordinate of the 1st ball and second ball as a functions of time would be given by...

Y1(t) = 40- 0.5gt^2 metres

Y2(t) = 20t + 0.5(-g)t^2 = 20t - 0.5gt^2 metres

where g = 9.8ms^-2 and t is in seconds

they collide then Y1 = Y2

which gives, at t = 2 they collide....

i.e., at a height of (40 - 0.5*9.8*4 ) metres

= 20.4 metres from ground.

2)

assume that at t = 0, the first stone is thrown....

Y1 = 20t - 0.5gt^2 metres

Y2 = 30t - 0.5g(t-1)^2 metres



they collide at the time 't' given by:

20t - 0.5gt^2 = 30t - 0.5g(t-1)^2

=> they collide at t = 0.2475 seconds.

=> they never collide....

.. as the second stone starts it's flight at t = 1second.