A markman fires a .22 caliber rifle horizontally at a target the bullet has a muzzle velocity with magnitude 750 ft/s. How much does the bullet drop in flight if the target is (a) 50.0 yd away and (b) 150.0 yd away?
for this question i was given the equation
S= ut + at^2/2
I don't know what numbers I should put where all i know is that a = -9.8 and u = 750 but the answer for the question is .64 how do they get this
You have the correct mathematical answer of -0.644.
But the question asks
"How much does the bullet drop in flight ..."
so the direction is implied, so the negative is not required.
You only have to translate the mathematical answer to a human answer.
Sorry Sarah if the previous explanation was not clear.
Since he shoots horizontally, the drop starts the moment the bullet leaves the rifle, at the rate of g (acceleration due to gravity).
For (a) and (b), calculate the time t the bullet is in the air using the horizontal speed, then calculate drop using
S=ut+at^2/2
where a=-g, and u the muzzle velocity.
The formula for S (distance) applies independently to horizontal and vertical directions.
u=initial velocity
a=acceleration
t=time
For part (a),
Horizontal direction:
time to reach 50 yards is the distance divided by the muzzle velocity. There is no acceleration (a=0) and air resistance is neglected.
S=ut + a t2/2
50*3 = 750 * t + 0
Thus
t=50 yards * 3 ft/yd / 750 ft/s
=0.2 s.
Vertical direction:
u=initial vertical velocity = 0 (bullet was shot horizontally)
a=-g = -32.2 ft/s/s
S = 0*t + a t2/2
= 0 + (-32.2)0.22/2 ft.
= 0 + (-32.2)* 0.04 /2 ft.
= 0.644 ft.
I am sure you can do part (b) along the same lines. Post your answer if you want it checked.
how did you .644 to not be a negative ???
Correct, it should be negative, since the bullet drops downwards.
Thank you.
well in the book it says its not negative either
To solve this problem, we need to determine the vertical displacement or drop of the bullet in flight at the given distances. We'll start by calculating the time it takes for the bullet to travel each distance.
We can use the equation of motion:
S = ut + (1/2)at^2
Let's plug in the known values for each part of the equation:
For part (a) at a distance of 50.0 yd:
S = ? (this is what we want to find)
u = 750 ft/s
a = -32.2 ft/s^2 (considering downward as negative)
In this case, we need to convert the distance from yards to feet:
50.0 yd * 3 ft/yd = 150 ft
Now, let's rearrange the equation to solve for time (t):
S = ut + (1/2)at^2
150 ft = 750 ft/s * t - (1/2) * 32.2 ft/s^2 * t^2
Simplifying:
0 = -16.1 t^2 + 750t - 150
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -16.1, b = 750, and c = -150.
t = (-750 ± √(750^2 - 4 * -16.1 * -150)) / (2 * -16.1)
t ≈ 0.918 s or t ≈ 46.8 s (ignoring the negative solution)
Now that we know the time it takes for the bullet to travel the distance of 50.0 yd, we can calculate the vertical displacement or bullet drop.
Using the equation:
S = ut + (1/2)at^2
S = -32.2 ft/s^2 * (0.918 s)^2 / 2 ≈ -13.7 ft
The negative sign indicates the bullet drops downwards. To convert this to yards, we divide by 3:
-13.7 ft / 3 ft/yd ≈ -4.6 yd
So, at a distance of 50.0 yd, the bullet drops approximately 4.6 yd downwards.
Now, let's move on to part (b) at a distance of 150.0 yd:
Follow the same steps as above, plugging in the new distance:
150.0 yd * 3 ft/yd = 450 ft
Using the quadratic formula, find t ≈ 2.75 s (ignoring the negative solution)
Using the equation: S = ut + (1/2)at^2
S = -32.2 ft/s^2 * (2.75 s)^2 / 2 ≈ -119.6 ft
Converting to yards: -119.6 ft / 3 ft/yd ≈ -39.9 yd
So, at a distance of 150.0 yd, the bullet drops approximately 39.9 yd downwards.
To summarize:
(a) At a distance of 50.0 yd, the bullet drops approximately 4.6 yd downwards.
(b) At a distance of 150.0 yd, the bullet drops approximately 39.9 yd downwards.
The given answer of 0.64 might be rounded to one decimal place, but based on the calculations, it seems to be inaccurate.