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?
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.
what is U? and i don't have time so i need to find it and what do i put in for S to find t? sorry im so confused
u is the initial velocity = muzzle velocity.
The formula is to find S, the distance the bullet drops vertically.
so u would = 750 ft/s, but there is no time so how can you solve for S if t is not available
The time is simply the horizontal distance divided by the muzzle speed, since the rifle is aimed horizontally, so cos0=1.
To find the amount the bullet drops during its flight, we need to consider the effects of gravity. The key concept to understand is that the bullet's path is a parabolic curve due to the combined effect of its horizontal velocity and the constant force of gravity pulling it downwards.
Let's start by calculating the time of flight (t) for each scenario using the formula:
t = d / v
where:
t = time of flight
d = distance to the target
v = muzzle velocity
(a) For a target distance of 50.0 yd:
t = 50.0 yd / 750 ft/s
To calculate the time in seconds, we need to convert yards to feet since the muzzle velocity is given in feet per second. There are 3 feet in a yard, so:
t = (50.0 yd * 3 ft/yd) / 750 ft/s
Now we can compute the time:
t ≈ 0.2 s
(b) For a target distance of 150.0 yd:
t = 150.0 yd / 750 ft/s
Converting yards to feet:
t = (150.0 yd * 3 ft/yd) / 750 ft/s
Calculating the time:
t ≈ 0.6 s
Now, to find the bullet drop during the time of flight, we can use the formula:
d_drop = 1/2 * g * t^2
where:
d_drop = bullet drop
g = acceleration due to gravity (32.2 ft/s^2)
t = time of flight
(a) For a target distance of 50.0 yd:
d_drop = 1/2 * 32.2 ft/s^2 * (0.2 s)^2
Calculating the bullet drop:
d_drop ≈ 0.64 ft
(b) For a target distance of 150.0 yd:
d_drop = 1/2 * 32.2 ft/s^2 * (0.6 s)^2
Calculating the bullet drop:
d_drop ≈ 5.78 ft
Therefore, the bullet drop in flight is approximately 0.64 ft for a target distance of 50.0 yd and 5.78 ft for a target distance of 150.0 yd.