KINEMATICS: How Far Will a Fired Projectile Travel?

Materials needed: ( 1 ) Spring-loaded cannon; ( 2 ) ruler; ( 3 ) timer

The goal will be to determine how far a model cannon’s projectile will travel. First, we will theoretically determine the muzzle velocity ( v_m ) of our model cannon. Position the cannon horizontally upon a table. Measure the vertical distance ( y ) between the cannon and the floor’s surface. 

Projectiles fired from the table in the ( x ) direction will move with a constant velocity ( v_x ) in the ( x ) direction, but they accelerate under the influence of gravity in the ( y ) direction at a rate of ( g ) = 9.8 m / s^2.

The kinematic equation that relates distance and time in the vertical ( y ) direction is ( y ) = Vy_it + ½ gt^2 will allow us to calculate the time it takes for the projectile to move from the tabletop to the ground’s surface. Since the fired projectile will travel in the horizontal and vertical directions simultaneously, this value for time is the total time ( t_t ) that the projectile spends in flight. Additionally, we measure the distance ( d ) that the projectile travels in the ( x ) direction. Since distance ( d ) = vt, where ( v ) = velocity, and ( t ) = time, ( d / t ) = muzzle velocity ( v_m ) of the projectile.

NOTE: If we choose to designate the downward direction as positive, acceleration due to gravity ( g ) will be additive with the falling projectile. Since ( g ) is additive, the form of ( y ) = Vy_it +/- ½ gt^2 used is ( y ) = Vy_it + ½ gt^2. 

Upon being fired, the projectile has an initial velocity ( v_x ) in the ( x ) direction, but its initial velocity in the ( y ) direction is Vy = 0, and ( y ) = 0 + ½ gt^2. Multiplying both sides of the equation by 2 gives us ( 2y ) = gt^2. Dividing both sides by ( g ) gives us 2y/g  = t^2. Finally, √ 2y/g = t.

NOTE: Compare the theoretical time of flight to the time that is measured with a stopwatch.

With the cannon in position upon the table top, fire the cannon in the horizontal direction. Allow the surface below to easily be marked or indented by the projectile when it lands ( Ex. sand ). Upon landing, measure how far the projectile traveled in the ( x ) direction. Take a couple of measurements and calculate the average distance ( d ) traveled.

We may now calculate the muzzle velocity ( v_m ) of the cannon. Since distance ( d ) = vt, ( v_m ) = ( d/t ) = [x/√ (2y/g)].

NOTE: Estimate ( v_m ) by dividing the value of ( x ) with the time estimated using a stopwatch, and compare this value of muzzle velocity with the value obtained above.

Let’s now determine how far the projectile will travel after being fired from the floor at some arbitrary angle. Since distance in the ( x ) direction = v_xt_t ( t_t = total time of flight ), we will first calculate how long it takes the projectile to reach its apex. When the projectile reaches its apex, half the total time of flight will have passed. Multiplying the value obtained by 2 will give us the total time of flight. Additionally, v_x = v_mcos θ. 

Confirm with a drawing that sin θ in terms of the projectile’s velocity will be (v_y/v_m), and thus, v_{y initial} = v_m sin θ. We don’t know how high the projectile will travel, so we must use an equation that relates velocity and time to determine how long it takes the projectile to reach its apex. When the projectile reaches its apex, v_{y final} = 0. Since v_{y final} = [v_{y initial} – gt] relates final velocity and time, 0 = v_{y initial} – gt. A negative value of ( g ) is used in the equation, because the force of gravity will cause the projectile to decelerate as it travels upward. Since v_{y initial} = v_msin θ, our calculation is as follows:

0 = v_msin θ – gt, and ( t ) = (-v_msin θ/-g ).  Total time of flight = ( 2 )( -v_msin θ/-g ).Finally, use a drawing to confirm that cos θ = v_x/v_m, and v_x = v_mcos θ. Since distance ( d ) = ( v )( t ), our projectile should travel a distance in the x-direction = ( Vx )( t total ) = [ ( v_mcos θ) ][ ( 2 )( v_msin θ ) /g ) ]