Question

Two projectiles are thrown from point A & B, towards each other, A and B are separated by 130 m. A is thrown with 20 m/s. Calculate the minimum velocity in m/s with which B is to be thrown, such that they just meet on the ground? (the projectiles can be thrown at any angle) [take g = 10 m/$s^2$]

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Answer 1

For both the projectiles, the angle of projection must be equal to ${45}^{\circ }$ so as to reach maximum range.

x+y = 130

For projectile A,

$x=\frac{u_A^2}{g}sin{90}^{\circ }=\frac{20\times 20}{10}=40$

So, y=130-40 = 90

Now, for the projectile B,

$90=\frac{u_B^{2}sin90}{g}$

$u_B^2=900\Rightarrow u_B=\pm 30$

Hence, x=30$\frac{m}{s}$ which is the velocity of B required for two projectiles to meet.