A 5.00 g bullet (bullet 1) is fired from a gun at height of 2.00 m above a horizontal surface. When the bullet
leaves the muzzle of the gun it is travelling at a velocity of <⃗# = 45.0 ms-1 2. At the same moment a similar ̂
is bullet (bullet 2) is dropped from rest from the same height. Ignore air resistance.
(a) Calculate the time it takes each bullet to reach the ground. Compare your answers.
(b) Draw a free body diagram for each bullet during the motion.
(c) Calculate the velocity with which each bullet hits the ground.
(d) How far down range does bullet 1 travel?
(e) If bullet 1 was fired a +45° angle (in reference to the +"-axis) from the same initial height, how far
would it have travelled down range when it hit the ground
(a) Well, bullet 1 seems to be in quite a hurry, while bullet 2 is taking its sweet time. To calculate the time it takes each bullet to reach the ground, we can use the kinematic equation:
Δy = v0y * t + (1/2) * g * t^2,
where Δy is the vertical displacement, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
For bullet 1:
Δy = -2.00 m (negative because it's moving downwards)
v0y = 45.0 m/s (since it was fired from the gun)
g = -9.81 m/s^2 (negative because it's directed downwards)
Plugging the values into the equation:
-2.00 m = (45.0 m/s) * t + (1/2) * (-9.81 m/s^2) * t^2.
Solving this quadratic equation will give us the time it takes for bullet 1 to reach the ground. On the other hand, for bullet 2, since it was dropped from rest, its initial vertical velocity v0y is zero. Thus, we can simplify the equation to:
-2.00 m = (1/2) * (-9.81 m/s^2) * t^2.
Again, solving for t will give us the time it takes for bullet 2 to reach the ground.
(b) Drawing a free body diagram for each bullet during the motion is a great idea. However, since I'm just a bot and don't have visual capabilities, I'll leave that to your imagination.
(c) To calculate the velocity with which each bullet hits the ground, we can use the formula:
vf = v0 + g * t,
where vf is the final velocity, v0 is the initial velocity, g is the acceleration due to gravity, and t is the time.
For bullet 1, since its initial velocity v0 is 45.0 m/s and it takes some time to reach the ground, we can substitute the value of t we calculated earlier to find the final velocity. As for bullet 2, since it was dropped from rest, its initial velocity v0 is zero. Therefore, its final velocity will simply be -g * t.
(d) To determine how far down range bullet 1 travels, we need to consider its horizontal motion. Without any information regarding the horizontal velocity or time, it's difficult to provide a specific answer. If you could provide more details, I'd be happy to help with the calculation.
(e) Ah, bullet 1 wants to take a detour and travel at a +45° angle? Well, that's a whole different story! To calculate how far it would have traveled down range when it hits the ground, we need to consider both its horizontal and vertical components of motion.
Again, without knowing the initial horizontal velocity or time of flight, it's impossible to give an exact answer. If you have any additional information, I'll be here, ready to assist you on this comedic journey!
(a) To calculate the time it takes for each bullet to reach the ground, we can use the equations of motion. The equation we will use is:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
For bullet 1:
h1 = 2.00 m
u1 = 45.0 m/s
g = 9.8 m/s^2
Substituting these values into the equation, we get:
2.00 = 45.0t + (1/2)(9.8)t^2
Simplifying the equation and set it equal to zero:
4.9t^2 + 45.0t - 2.00 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. By using the quadratic formula, we get:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-45.0 ± √(45.0^2 - 4*4.9*(-2.00))) / (2*4.9)
Solving for t, we get two solutions: t ≈ 0.139 s and t ≈ -9.724 s. Since time cannot be negative, the only valid solution is t ≈ 0.139 s.
So, bullet 1 takes approximately 0.139 seconds to reach the ground.
For bullet 2:
h2 = 2.00 m
u2 = 0 m/s (dropped from rest)
g = 9.8 m/s^2
Using the same equation, we have:
2.00 = 0t + (1/2)(9.8)t^2
Simplifying the equation:
4.9t^2 = 2.00
Solving for t, we get:
t ≈ √(2.00/4.9) ≈ 0.202 s
So, bullet 2 takes approximately 0.202 seconds to reach the ground.
Comparing the answers, we see that bullet 1 reaches the ground faster than bullet 2.
(b) The free body diagram for each bullet during motion will show the gravitational force acting downwards, the weight of the bullet, and the normal force. Since both bullets are falling downward, the only force acting on them is gravity.
Bullet 1:
--> Gravitational force
|
v
_ _ _ _
Bullet 2:
--> Gravitational force
|
v
_ _ _ _
(c) To calculate the velocity with which each bullet hits the ground, we can use the equation:
v = u + gt
For bullet 1:
u1 = 45.0 m/s
t = 0.139 s
g = 9.8 m/s^2
Substituting these values into the equation, we have:
v1 = 45.0 + (9.8)(0.139) ≈ 46.36 m/s
So, bullet 1 hits the ground with a velocity of approximately 46.36 m/s.
For bullet 2:
u2 = 0 m/s
t = 0.202 s
g = 9.8 m/s^2
Using the equation, we have:
v2 = 0 + (9.8)(0.202) ≈ 1.99 m/s
So, bullet 2 hits the ground with a velocity of approximately 1.99 m/s.
(d) To determine how far downrange bullet 1 travels, we can use the equation of motion:
s = ut + (1/2)gt^2
where s is the distance traveled.
For bullet 1:
u1 = 45.0 m/s
t = 0.139 s
g = 9.8 m/s^2
Substituting these values, we have:
s1 = (45.0)(0.139) + (1/2)(9.8)(0.139)^2 ≈ 3.035 m
So, bullet 1 travels approximately 3.035 meters downrange.
(e) If bullet 1 was fired at a +45° angle from the same initial height, we need to split its initial velocity into horizontal and vertical components. The horizontal component is u1x = u1 * cos(45°) and the vertical component is u1y = u1 * sin(45°).
Since the height is the same, we can calculate the time it takes for bullet 1 to reach the ground using the vertical component of velocity (u1y) and the equation of motion:
h = uyt + (1/2)gt^2
where h is the height, uy is the vertical component of velocity, g is the acceleration due to gravity, and t is the time.
For bullet 1:
h = 2.00 m
uy = u1 * sin(45°)
g = 9.8 m/s^2
Doing the calculation, we have:
2.00 = (45.0 * sin(45°))t + (1/2)(9.8)t^2
Simplifying the equation, we have:
4.9t^2 + (45.0 * sin(45°))t - 2.00 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-(45.0 * sin(45°)) ± √((45.0 * sin(45°))^2 - 4 * 4.9 * (-2.00))) / (2 * 4.9)
Solving for t, we get two solutions: t ≈ 0.206 s and t ≈ -3.807 s. Since time cannot be negative, the valid solution is t ≈ 0.206 s.
To find the horizontal distance traveled (downrange), we can use the horizontal component of velocity (u1x) and the time (t):
s = u1x * t
For bullet 1:
u1x = u1 * cos(45°)
t = 0.206 s
Substituting the values, we have:
s = (45.0 * cos(45°)) * 0.206 ≈ 15.35 m
So, if bullet 1 was fired at a +45° angle, it would travel approximately 15.35 meters downrange when it hits the ground.
To answer these questions, we need to use the equations of motion and the principles of projectile motion. Let's go step by step to solve each part of the problem.
(a) To calculate the time each bullet takes to reach the ground, we can use the equation for vertical displacement:
Δy = v₀y * t + 0.5 * a * t²
Where
Δy is the vertical displacement (in this case, it's -2.00 m since the bullet is moving downward),
v₀y is the initial vertical velocity,
a is the acceleration in the vertical direction (in this case, acceleration due to gravity, which is -9.8 m/s²),
t is the time.
For bullet 1:
Δy₁ = -2.00 m,
v₀y₁ = 45.0 m/s,
a = -9.8 m/s².
Plugging in the values and rearranging the equation, we can solve for t₁.
-2.00 m = (45.0 m/s) * t₁ + 0.5 * (-9.8 m/s²) * t₁²
This is a quadratic equation, so by solving it, we can find the time t₁.
For bullet 2:
Δy₂ = -2.00 m,
v₀y₂ = 0 m/s,
a = -9.8 m/s².
Using the same equation but with different initial velocity, we can solve for t₂.
(b) To draw a free body diagram for each bullet during motion, we need to consider the forces acting on them. In this case, the only significant force is the gravitational force, which always acts downward.
For bullet 1 in motion, the gravitational force is acting vertically downward.
For bullet 2, since it is dropped from rest, the only force acting on it is the gravitational force vertically downward.
(c) To calculate the velocity with which each bullet hits the ground, we can use the equation for final velocity:
v = v₀ + a * t
For bullet 1, we know the initial velocity v₀₁ and time t₁, so we can calculate the final velocity v₁.
For bullet 2, the initial velocity v₀₂ is 0 m/s, and we know the time t₂, so we can calculate the final velocity v₂.
(d) To determine how far downrange bullet 1 travels, we need to calculate the horizontal distance traveled. Since there is no horizontal force acting on bullet 1, its horizontal velocity remains constant.
We can use the equation for horizontal displacement:
Δx = v₀x * t
where
Δx is the horizontal displacement (downrange distance),
v₀x is the initial horizontal velocity (equal to v₀₁ since there is no acceleration in the horizontal direction),
t is the time calculated earlier.
(e) If bullet 1 was fired at a +45° angle from the same initial height, we need to calculate the horizontal distance it would have traveled when it hit the ground. In this case, we need to find the horizontal component of the initial velocity, which is v₀x.
Given the launch angle of 45°, the vertical component of velocity is v₀y₁ = v₀ * sin(45°), and the horizontal component of velocity is v₀x = v₀ * cos(45°).
Now, we can use the same equation for horizontal displacement as before to calculate the distance traveled.