Question

(1) A short find honontally at a velocity of 400 m/s. Find the magnitude and direction of its velocity after 10 sec.​

Answer 1

Answer:

412.31 m/s in a direction making angle θ = $tan^{-1} (\frac{1}{4})$ with the horizontal downwards i.e., ≈ 14.04° .

Explanation:

First of all, according to the equation

$s = ut + \frac{1}{2} gt^{2}$

the gun, which is firing the bullet, must be 500 meters above the ground to keep the bullet in air for 10 seconds. ( taking g = 10 $m/s^{2}$ )

If the gun is at appropriate height, then after 10 seconds, it will have a vertical down velocity of 100 m/s, and horizontal velocity of 400 m/s. (Neglecting the air resistance)

So vector sum will be 412.31 m/s in a direction downwards with horizontal making angle θ = $tan^{-1} (\frac{1}{4})$ .