A horizontal rifle is fired at a target 47 m away. The bullet falls vertically a distance of 8 cm on the way to the target.
(a) What is the speed of the bullet just after it left the rifle?
(b) What is the speed of the bullet just before it hits the target?
L =47 m, Δy = h= 8 cm=0.08 m. V(x) = ?
h =g•t²/2, t = sqrt(2•h/g) =
= sqrt(2•0.08/9.8) =0.1278 s.
L =V(x) •t = > V(x) =L/t =
= 47/0.1278 =367.76 m/s.
V(y) - g•t = 9.8•0.1278 = 1.252 m/s.
V= sqrt[V(x)² + V(y)²] =
=sqrt[(367.76)²+ (1.252)²] =367.77 m/s.
(a) Well, let me calculate that for you! To find the speed of the bullet just after it left the rifle, we need to consider the vertical distance it fell. So, since it fell 8 cm vertically, we can use that information to determine the time it took to reach the target. Using the equation for free fall, we can say that the distance fallen is equal to half the acceleration due to gravity times the square of the time. In this case, 8 cm is equal to half of 9.8 m/s² times the square of the time. Solving for time, we can find that it took about 0.090 seconds for the bullet to reach the target after leaving the rifle. Now, we can use this time to calculate the initial velocity of the bullet by dividing the horizontal distance of 47 m by the time of 0.090 seconds. So, the speed of the bullet just after it left the rifle is roughly 522 m/s.
(b) To find the speed of the bullet just before it hits the target, we can use the horizontal distance and the time it took to reach the target. We already know that it took about 0.090 seconds for the bullet to reach the target, and the horizontal distance is 47 m. So, the speed of the bullet just before it hits the target can be calculated by dividing the horizontal distance by the time. This gives us a speed of approximately 522 m/s.
So, for both questions, the speed of the bullet is approximately 522 m/s. Keep in mind, these calculations are just rough estimates and don't take into account other factors that may affect the bullet's trajectory, such as air resistance or the specific type of bullet used.
To find the answer to these questions, we will need to use some equations of motion and principles of projectile motion. Let's break the problem down step by step:
(a) What is the speed of the bullet just after it left the rifle?
To determine the initial speed (also known as the muzzle velocity) of the bullet just after it left the rifle, we can use the principle of projectile motion that states the horizontal and vertical motions are independent of each other. This means that the initial vertical velocity is only affected by the force of gravity, while the horizontal velocity remains constant.
Since the bullet falls vertically a distance of 8 cm on its way to the target, we can conclude that it took some time for it to reach the target. Let's call this time "t."
To find the initial vertical velocity, we can use the formula for vertical displacement:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy = vertical displacement (in this case, 8 cm = 0.08 m)
v₀y = initial vertical velocity (which is what we're trying to find)
a = acceleration due to gravity (which is approximately 9.8 m/s²)
t = time taken to reach the target (which is the same as the time of flight)
Substituting the given values into the equation, we get:
0.08 m = v₀y * t - (1/2) * 9.8 m/s² * t²
Now, we need to find the time of flight. For a projectile launched horizontally, the time of flight can be determined using the horizontal distance traveled and the horizontal velocity, based on the equation:
d = v₀x * t
Where:
d = horizontal distance traveled (47 m in this case)
v₀x = initial horizontal velocity (which is what we're trying to find)
t = time of flight
Rearranging the formula, we find:
t = d / v₀x
Substituting this value of "t" into the earlier equation, we have:
0.08 m = v₀y * (d / v₀x) - (1/2) * 9.8 m/s² * (d / v₀x)²
Simplifying the equation, we get:
0.08 m = (d * v₀y) / v₀x - (1/2) * 9.8 m/s² * (d² / v₀x²)
We can rearrange this equation to isolate v₀x:
0.08 m * v₀x² = d * v₀y - (1/2) * 9.8 m/s² * d²
Simplifying further:
0.08 m * v₀x² + (1/2) * 9.8 m/s² * d² = d * v₀y
Now, let's solve for v₀x:
v₀x = sqrt((d * v₀y) / (0.08 m) + (1/2) * 9.8 m/s² * d² / (0.08 m))
Plugging in the given values, we get:
v₀x = sqrt((47 m * v₀y) / (0.08 m) + (1/2) * 9.8 m/s² * (47 m)² / (0.08 m))
Now we need to determine the value of v₀y. Since the bullet is fired horizontally, the initial vertical velocity (v₀y) is zero. Therefore, the formula simplifies to:
v₀x = sqrt((1/2) * 9.8 m/s² * (47 m)² / (0.08 m))
Evaluating this expression, we find:
v₀x ≈ 614.67 m/s
So, the speed of the bullet just after it left the rifle is approximately 614.67 m/s.
(b) What is the speed of the bullet just before it hits the target?
Since there are no external horizontal forces acting on the bullet during its flight, its horizontal speed remains constant. Therefore, the speed of the bullet just before it hits the target is the same as its initial horizontal speed, which we found to be approximately 614.67 m/s.
Therefore, the speed of the bullet just before it hits the target is approximately 614.67 m/s.