Question

A particle P is projected from a point O with an initial velocity of 70m/s at an angle 30° to the horizontal. At the same instant a second particle Q is projected in the same direction, 20m from point O with initial speed 25m/s from a point level with O. If Particle Q is displaced to 90m from point of release C. If the particles Q collide with P at 80m from the point of release C, find the angle of projection of Q and find when the collision occurs.

Answer 1

for  projectile:

$x=x_0+v_{0x}t$

$y=y_0+v_{0y}t-gt^2/2$

for particle P:

$x_P=70tcos30\degree$

$y_P=70tsin30\degree-4.9t^2$

for particle Q:

$x_Q=25tcos\alpha+20$

$y_Q=25tsin\alpha-4.9t^2$

If the particles Q collide with P, then:

$y_P=y_Q$

$70sin30\degree=25sin\alpha$

$sin\alpha=70sin30\degree/25=7/5>1$

then:

$y_Q=25tsin\alpha+4.9t^2$ (Q is projected in -y direction)

for collision:

$70tsin30\degree-4.9t^2=25tsin\alpha+4.9t^2$

$70sin30\degree-9.8t=25sin\alpha$

$x_Q^2+y_Q^2=(25tsin\alpha+4.9t^2)^2+(25tcos\alpha+20)^2=80^2$

$625t^2+9.8t^3(35-9.8t)+24t^4+1000tcos\alpha+400=6400$

$x_P=x_Q$

$25tcos\alpha+20=70tcos30\degree$

then:

$625t^2+9.8t^3(35-9.8t)+24t^4+40(35t\sqrt 3-20)=6000$

$-72t^4+343t^3+625t^2+2425t-6800=0$

time of collision:

$t_1=1.66$ s, $t_2=6.54$ s

$sin\alpha_1=(35-9.8\cdot1.66)/25=0.749$

$\alpha_1=48.5\degree$

$sin\alpha_1=(35-9.8\cdot6.54)/25=-1.16<-1$

so, we have:

angle of projection of Q:

$\alpha_1=-48.5\degree$ (since Q is projected in -y direction)

time of collision:

$t=1.66$ s