A bullet is fired from a rifle that is held 4.6 m above the ground in a horizontal position. The initial speed of the bullet is 1130 m/s. There is no error in the velocity value.
(a) Find the time it takes for the bullet to strike the ground.
(b) Find the horizontal distance traveled by the bullet.
(c) If the error in time measurements is Δt, what will be the equation to calculate error in the horizontal distance (Δx)? (Assume x is in meters and t is in seconds. Do not enter units in your expression. Substitute numeric values; the only variable you should enter is Δt.)
Plug the values you have into the formula y = y[initial] + 1/2gt^2
Remember 1/2g is a negative value!
y[initial] is going to be the height the bullet starts from, or rather the height at which the gun is held. g is the standard gravitational acceleration. Y will just be zero, since the final height of the bullet will be the ground, or 0.
Next, solve! Subtract y[initial] from y, and then divide by 1/2g. This will equal t^2. Take the square root and viola!
For the second part of the question, use the distance formula: d = vt + 1/2at^2.
Once again, a will be the gravitational acceleration, and negative!
Not sure about the last one....
Hope it helps!
This is wrong!
This is right, just plug the numbers in
To find the answers to these questions, we can use the equations of motion and the known values:
(a) Find the time it takes for the bullet to strike the ground:
We can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
h = initial height = 4.6 m
u = initial vertical velocity = 0 m/s (since the bullet is fired horizontally)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time taken to reach the ground (which we need to find)
Rearranging the equation, we have:
(1/2)gt^2 = h
(1/2)(9.8)t^2 = 4.6
4.9t^2 = 4.6
t^2 = 4.6/4.9
t^2 = 0.9387755
t ≈ √0.9387755
t ≈ 0.968 s (approx.)
Therefore, the time it takes for the bullet to strike the ground is approximately 0.968 seconds.
(b) Find the horizontal distance traveled by the bullet:
We can use the equation of motion for horizontal motion:
x = ut
Where:
x = horizontal distance traveled
u = initial horizontal velocity = 1130 m/s
t = time taken to reach the ground (which we found in part a)
Plugging in the values we know:
x = (1130 m/s)(0.968 s)
x ≈ 1094.24 m (approx.)
Therefore, the horizontal distance traveled by the bullet is approximately 1094.24 meters.
(c) If the error in time measurements is Δt, the equation to calculate error in the horizontal distance (Δx) can be found using the chain rule of differentiation. Since the equation for horizontal distance is x = ut, we can differentiate it with respect to time (t) to find the rate of change of x with respect to t, which is the velocity (v).
Therefore, Δx = v Δt
Since v is constant in this case (1130 m/s), the equation simplifies to:
Δx = 1130 Δt
Hence, the equation to calculate the error in the horizontal distance (Δx) is 1130 times the error in time measurements (Δt).