Question

Using only the spring gun with the catcher removed, fire the steel ball, making sure the launch velocity is horizontal. Using carbon paper (and something to catch the ball after impact with the floor), determine the ball’s horizontal range and its total vertical displacement (i.e., height at point of firing). Repeat several times in order to obtain statistical data.

Derive a formula in symbolic form for the initial velocity of the projectile in terms of the following three quantities only: the horizontal range (x), the vertical height at launch (y) and the acceleration due to gravity (g). Finally, determine the initial velocity of the steel ball and refer to this velocity as v1 .

Need help deriving the equation (italics). Please show all intermediate derivations too (i.e. how to come to horizontal range (x) = v1*t). Thank you!

Answer 1

As given in the data but initial velocity is only along horizontal.

I.e. initially there is no vertical component of velocity

This is the case of motion in a plane.

Motion must be studied in two dimensions

I.e.

In X direction	in Y direction
Displacement= x	displacement=y
Intial Horizontal velocity=vx= v1	intial vertical component of velocity=0
Acceleration= 0 Using S= ut+1/2(at^2) in horizontal direction S- Displacement V final velocity ,u initial velocity, t time X = v1 t t= X/v1 Here X is the horizontal range Hence X = V1(√2y/g)	acceleration= -g = acceleration due to gravity= -9.8m/s^2 Using same formula S= ut+1/2(at^2) in vertical direction Y= 0(t)+1/2(-g) (t)^2 But t= X/V1 Y= -1/2 g(x/v1)^2

Hence initial velocity v1 = x/√(2y/g)

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