Projectile Motion Question (1 Viewer)

[Yell0w]

A ball is dropped from a looks fired out 320m high. At the same time, a stone is fired vertically upwards from the valley floor with a speed of V m/s.
(Taking g = 10m/s)

i. Show that if the ball and the stone collided in the air, then they do so when V >= 40m/s

ii. Find, in terms of V, the height when a collision occurs.

Thanks.
~Yell0w

 

[k02033]

Let Y1(t) be height of the ball at time t

Let Y2(t) by height of stone at time t

Y1"=-g Where g=10ms^-2

so Y1'(t)=-gt+C, but since ball is dropped Y1'(0)=0 which implies c=0

so Y1'(t)=-gt

so Y1(t)=-gt^2+K, but since at t=0 ball was at the top of the cliff so Y1(0)=320 so K=320

so Y1(t)=-gt^2+320.

solve for Y1(t)=0 gives us t=8,-8, which means the ball can only stay in the air for 8 secs

Now similar method and argument gives us Y2(t)=-gt^2+Vt

equate Y1(t) and Y2(t), ie we are trying to find when they collide, gives Vt=320

so t=320/V is time of collision

but t needs to be < or = 8 since the ball can only stay in the air for 8 secs

so V> or=40ms^-1

2) collision occurs at t=320/V so sub it into Y1(t) or Y2(t) to get collision height

 

[Yell0w]

Thanks Alot =]. Rep++