Answer to Question #266720 in Physics for Nishat

Question

A projectile is fired at an angle 15 ° from a gun that is 45 m above the flat ground, emerging

from the gun with a speed of 300 m/s.

(a) How long does the projectile remain in air?

(b) At what horizontal distance from the firing ground does it strike the ground.

(c) What is the velocity of the projectile when it hits the ground?

Answer 1

(a) How long the projectile remains in air:

$$T=\frac{v\sin\theta}{g}+\sqrt{\frac{2[h_0+(v\sin\theta)^2/(2g)]}{g}}=16.4\text{ s}.$$

(b) The horizontal distance from the firing point to the point where it strikes the ground:

$$R=Tv\cos\theta=4752\text{ m}.$$

(c) The velocity of the projectile when it hits the ground:

$$u=\sqrt{[v\cos\theta]^2+\Big[\sqrt{2g(h_0+[(v\sin\theta)^2/(2g)]}\Big]^2}=\\=301\text{ m/s}.$$