A paintball is fired from a gun horizontally with a speed of 110 m/s. If it hits a point 14 cm below the target:
A) What is the total time the paintball is in the air?
B) How far is the target from the paintball?
Simple projectile motion. Draw a picture.
yu have to figure out how long it takes to fall 14 cm
.14 = 4.9t^2
t = 0.169 sec
so, at 110 m/s, the ball travels horizontally 110*.169 = 18.6 m
To find the total time the paintball is in the air and the distance to the target, we can use the equations of motion for horizontally launched projectiles.
First, let's calculate the time of flight (A):
We know that the paintball is fired horizontally, which means there is no initial vertical velocity. Therefore, the time of flight will be the same as the time it takes for the paintball to fall down 14 cm (0.14 m).
1. Use the equation: Δy = Vyi * t + 0.5 * g * t^2
Since there is no initial vertical velocity (Vyi = 0), the equation becomes: Δy = 0.5 * g * t^2
Rearrange the equation to solve for time: t = sqrt(2 * Δy / g)
Plug in the values:
Δy = 0.14 m
g ≈ 9.8 m/s² (acceleration due to gravity)
t = sqrt(2 * 0.14 / 9.8)
t ≈ 0.165 s
Therefore, the total time the paintball is in the air is approximately 0.165 seconds.
Next, let's calculate the horizontal distance to the target (B):
2. Use the equation: Δx = Vxi * t
Since the paintball is fired horizontally, there is no initial horizontal velocity (Vxi = 0).
So, the equation simplifies to: Δx = 0 * t
As a result, the horizontal distance to the target (Δx) is 0 meters.
Since the paintball is fired horizontally, it reaches the target vertically but does not travel horizontally.