2 projectiles are fired (Ma= 3 kg, Mb= 2 kg) simultaneously from locations shown

Question

2 projectiles are fired (Ma= 3 kg, Mb=   2 kg) simultaneously from locations shown on the ground(Distance from A to B is, left to right is d=35m) at a angle of alpha=45°(for Vao) and Beta=60°(for Vbo). They are fired with initial velocities of Vao=25 m/s and Vbo= ?.

a) Find the launch speed Vbo of B so that the two projectiles will actually collide somewhere(This means they must reach that point at the same time t)

b) For this Vbo, assuming they stick together after impact, find where the combined projectile eventually hits the ground. Express the answer as a distance to the left or right of the launch point of A.

Explanation / Answer

Ma = 3 kg,
Mb = 2 kg

Horizontal velocity of projectile A, Vah = 25 * cos(45) = 17.67 m/s
Vertical velocity of projectile A, Vay = 25 * sin(45) = 17.67 m/s

Horizontal velocity of projectile B, Vbh = Vb * cos(60)
Vertical velocity of projectile B, Vby = Vb * sin(60)

Let's say they meet after time t,  
Horizontal Distance covered by projectile A, = Vah * t
Horizontal Distance covered by projectile B, = Vbh * t

Vah + Vbh = 35
17.67 * t + Vb * cos(60) = 35 ---------1

Considering Vertical distance,
Vay * t - 1/2*9.8*t^2 = Vby * t - 1/2*9.8*t^2
Vay = Vby
17.67 = Vb * sin(60)
Vb = 20.4 m/s

17.67 * t + 20.4 * cos(60) = 35
t = 1.40 s

sy = Vay * t - 1/2*9.8*t^2
sy = 17.67 * 1.4 - 1/2 * 9.8 * 1.4^2
sy = 15.13 m

vaf = 17.67 - 9.8*1.4
vaf = 3.95 m/s

Using Momentum conservation,
Initial Momentum = Final Momentum
ma * vah - mb * vbh = (ma + mb) * vfh
3 * 17.67 - 2 * 20.4 * cos(60) = (2 + 3) * vfh
vfh = 6.5 m/s

ma * vaf + mb * vbf = (ma + mb) * vfy
3 * 3.95 + 2 * 3.95 = (2 + 3) * vfy
vfy = 3.95 m/s

s = u*t + 1/2*9.8*t^2
-15.13 = 3.95*t - 1/2*9.8*t^2
t = 2.2 s

Horizontal distance covered = 6.5 * 2.2 = 14.3 m

Final point where combined projectile eventually hits the ground, = 17.67 * 1.4 + 14.3
= 39.0 m to the Right of A

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