A student decides to measure the muzzle velocity of a pellet from his gun. He points the gun horizontally. He places a target on a vertical wall a distance 3.27 m away from the gun. The pellet hits the target a vertical distance 0.264 m below the gun. What is the speed of the pellet?

8.36 m/sec

Explanation:

Since muzzel velocity is required to be determined i-e velocity with which bullet leaves the gun. So we have to suppose that bullet has no acceleration and moves with constant velocity.

a=0 m/sec2

Vi=0 (initially the bullet is at rest)

S= 3.27 m

using m/s

S= Vit+1/2at2

==> 3.27 = 0+0.5×

==> = 6.54

==> t=2.56 sec

Now using S= V×t

==> V = S/t

==> V= 3.27 × 2.56 = 8.36 m/sec

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i don't understand. explain it clearly