Projectile A is shot at angle 37 degrees above the horizonta

671 Projectile A is shot at angle 37° above the Take g-10 m/see horizontal with initial velocity 100 m/sec. Projectile B is fired at the same instant from a point 600 meters to the right of and 300 meters above the firing point of A. B is shot horizontal in the vertical plane of A\'s motion. V- 100m/s. 300 m. 37° What must the initial velocity of B be in order to cause the two projectiles to collide? At what height, h, do they collide? Take g 10 m/s (instead of 9.8) to simplify the arithmetic. 600 m

Solution

Suppose both collisdes after time t.

So at time t, their x and y coordinate will be same. (for A and B)

after time t:

for A:

Ax = vx *t = 100cos37 * t = 79.86t

Ay = vy*t + ay*t^2/2

Ay = (100sin37 * t) - (10 * t^2 /2 ) = 60.18t - 5t^2

for B:

Bx = 600 - vt

By = 300 - (10t^2 /2) = 300 - 5t^2

Ay = By

60.18t - 5t^2 = 300 - 5t^2

t = 4.98 sec

and Ax = Bx

79.86t = 600 - vt

79.86(4.98) = 600 - v(4.98)

v = 40.54 m/s ..........Ans

height = Ay = By = 60.18(4.98) - 5(4.98) ^2 = 300 - 5(4.98)^2 = 238 m .....Ans