Question

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x away from the gun. The pellet hits the target a vertical distance y below the gun.
(a) Show that the position of the pellet when traveling through the air is given by y = Ax^2, where A is a constant.

(b) Express the constant A in terms of the initial velocity v and the free-fall acceleration g.

(c) If x = 3 m and y = 0.21 m, what is the initial speed of the pellet?

Answer 1

Answer:

a)  y = A x² , b)   A = - ½ g / v₀², c)    v₀ = 15.46 m / s

Explanation:

For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same

X axis

         x = v₀ₓ t

         t = x / v₀ₓ

Y Axis  

         y = $v_{oy}$ t - ½ g t²

As it shoots horizontally the vertical speed is zero

        y = - ½ g t²

We replace

         y = - ½ g (x / v₀ₓ)²

The initial speed is all horizontal

        v₀ₓ = v₀

         

         y = - ½ g / v₀²    x²

        y = A x²

b) the expression for the constant is

           A = - ½ g / v₀²

c) we look for the initial speed

           v₀² = - ½ g x² / y

   As the object falls below the exit point its height is negative

          v₀ = √ (- ½  9.8   3²/ (-0.21))

          v₀ = 15.46 m / s