The purpose of this investigation is to measure the vertical displacement, or height of the launch, and the horizontal displacement, or range, travelled by a projectile (bullet from toy gun).
What is the shape of the actual path travelled by a projectile? How closely does an actual projectile's results follow the theoretical predicted results?
The shape of the path travelled by the projectile, in this case the bullet of the gun, is a parabolic. This means that is a curvy shape due to the bullet being launched in the air (making curve go up) and the earth's gravity pulling it down (making curve go down). As the height of the bullet's release increases the the time to reach the ground will increase, and therefore the range of the bullet will increase. This is because the bullet's vertical velocity will decrease later as the height is higher up, having a larger time, and therefore a larger range.
A metre stick was used to measure the height and the range of the bullet. A stop was used to determine the time it took for the bullet to reach the ground. As the bullet was released, its path was parabolic. This means that its was curvy because it was first int air, but the gravity pulled it back down to the surface. The toy gun was steadily held in my hand.
The initial height was the distance from the gun to the surface used. The gun shot out the bullets at a fairly fast speed.
As the height was increased, the more time the bullet took to reach the ground. As the height was increased, the range was also higher.
Data of various heights used:
| Height (cm) |Time (seconds) |Horizontal Distance (cm) | |25 |2.26 |70.7 | |50 |3.19 |100 | |75 |3.91 |122 | |100 |4.52 |141 |
Picture of the launcher:
Height vs. Range graph- Refer to attached data in the back.
Position vs. Time graph- Refer to attached data in the back.
The graph results definitely support the hypothesis. This is because as the height of the toy gun was increased, the horizontal distance increased. Also, as the horizontal distance of the bullet increased, so did the time (vice- versa). The graphs were very similar due to the horizontal distance (cm) being constant on the y- axis of the graph. In the Horizontal Distance vs. Time graph, the time represented the corresponding heights of the Horizontal Distance vs Height graph. Making the graphs very similar.
Determining the Vi of the Bullet:
Vi = aav x Δt
aav = -9.81 m/s²
Δt = 3.19 seconds
Vi = -9.81 x 3.19
Vi = 31.3 m/s [↓]
*Therefore the initial velocity of the bullet is 31.3 m/s [↓].
Theoretical Ranges of the Bullet:
Formula- Δd = Vi x Δt
|Height (cm): |Range/ Horizontal Displacement (cm) : | |25 |Δd = 31.3 x 2.26 | | |Δd = 70.7 cm | |50 | Δd = 31.3 x 3.19 | | |Δd = 99.8 cm | |75 | Δd = 31.3 x 3.91 | |...