A military helicopter on a training mission is flying horizontally at a speed of 75.0m/s when it accidentally drops a bomb (fortunately, not armed) at an elevation of 310m . You can ignore air resistance.
a:How much time is required for the bomb to reach the earth?
b:How far does it travel horizontally while falling?
c:Find the horizontal component of the bomb's velocity just before it strikes the earth.
d:Find the vertical component of the bomb's velocity just before it strikes the earth.
e:If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?
a. h = 0.5g*t^2 = 310 m.
4.9t^2 = 310
t^2 = 63.27
Tf = 7.95 s. = Fall time.
b.
c. X = Xo = 75 m/s. Remains constant.
d. Y=Yo + g*Tf = 0 + 9.8*7.95=77.95 m/s
e. Dx = Xo*Tf = 75m/s * 7.95s. = 596 m.
Fom drop-off point.
a: How much time is required for the bomb to reach the earth?
Well, let's do some math! Since we don't have to worry about air resistance, we can use the basic formula for falling objects:
Distance = (Initial velocity x Time) + (1/2 x Acceleration x Time^2)
In this case, the initial velocity of the bomb is 0 m/s (since it's dropped), the acceleration is due to gravity (-9.8 m/s^2), and the distance it needs to fall is 310 m.
So plugging in the values, we get: 310 = (0 x Time) + (1/2 x -9.8 x Time^2)
Now we just need to solve for Time. But I think I'll leave that to you. After all, solving equations is just like defusing a bomb, both require a calm mind and steady hands!
b: How far does it travel horizontally while falling?
Since the helicopter is flying horizontally at a constant speed, the horizontal distance the bomb travels is the same as the horizontal distance traveled by the helicopter in the time it takes for the bomb to fall.
To find this distance, we can use the equation: Distance = Speed x Time.
We know the speed of the helicopter is 75.0 m/s, and we can find the time it takes for the bomb to reach the ground from part a. So, it's just a matter of plugging in the numbers and calculating the answer. Tada!
c: Find the horizontal component of the bomb's velocity just before it strikes the earth.
Since there are no horizontal forces acting on the bomb, its horizontal velocity remains constant throughout its fall. Therefore, the horizontal component of its velocity just before it strikes the earth is the same as the initial horizontal velocity of the helicopter, which is 75.0 m/s. So, it doesn't change!
d: Find the vertical component of the bomb's velocity just before it strikes the earth.
Well, just before the bomb hits the ground, its vertical velocity will be its final velocity in the vertical direction. We can calculate this using the equation: Final velocity = Initial velocity + (Acceleration x Time).
The initial velocity of the bomb is 0 m/s (since it's dropped), the acceleration is due to gravity (-9.8 m/s^2), and we can find the time it takes for the bomb to reach the ground from part a. Plug in those numbers, crunch some math, and you'll have your answer!
e: If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?
Well, since we know the horizontal distance traveled by the bomb (from part b) and the constant horizontal speed of the helicopter, we can conclude that the helicopter will be exactly above the spot where the bomb hits the ground. You could say it's a helicopter's way of playing "catch" with the bomb!
To solve this problem, we can use the basic equations of motion. Let's consider the following variables:
Initial vertical position (s0) = 310 m
Initial vertical velocity (v0y) = 0 m/s (since the bomb is dropped, it has no initial vertical velocity)
Vertical acceleration (ay) = 9.81 m/s^2 (acceleration due to gravity)
Horizontal velocity of the helicopter (vH) = 75.0 m/s
a) To find the time required for the bomb to reach the earth, we can use the equation for vertical displacement:
s = s0 + v0y * t + (1/2) * ay * t^2
Setting s = 0 (since the bomb hits the ground), s0 = 310 m, v0y = 0 m/s, and ay = 9.81 m/s^2, we can solve for t:
0 = 310 + 0 * t + (1/2) * 9.81 * t^2
This is a quadratic equation, so we can solve for t using the quadratic formula. The final equation will be:
0 = 4.905 * t^2
Solving for t gives us:
t = sqrt(0/4.905) = 0 s
Therefore, it takes 0 seconds for the bomb to reach the earth.
b) Since we know the horizontal velocity of the helicopter (75.0 m/s), we can calculate the horizontal distance traveled by the bomb using the equation:
Distance (d) = vH * t
Substituting the time t = 0 s, we find:
d = 75.0 * 0 = 0 m
Therefore, the bomb travels a horizontal distance of 0 meters while falling.
c) Just before the bomb strikes the earth, its horizontal velocity is still equal to the helicopter's horizontal velocity, which is 75.0 m/s.
d) Since there are no vertical forces acting on the bomb except gravity, its vertical velocity just before it strikes the earth will be equal to the velocity it gained from free fall. We can find it using the following equation:
vFy = v0y + ay * t
Substituting v0y = 0 m/s, ay = 9.81 m/s^2, and t = 0 s, we find:
vFy = 0 + 9.81 * 0 = 0 m/s
Therefore, the vertical component of the bomb's velocity just before it strikes the earth is 0 m/s.
e) Since the bomb takes 0 seconds to reach the ground, the helicopter would be directly above the point where the bomb is dropped.
To solve this problem, you can use the equations of motion. Let's break down each part of the question and explain how to find the answers:
a) How much time is required for the bomb to reach the earth?
To find the time required for the bomb to reach the ground, we can use the equation of motion for vertical motion:
s = ut + 0.5 * a * t^2
where s is the displacement (change in height), u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
In this case, the initial vertical velocity (u) is 0 m/s (as the bomb is not moving up or down at the moment it is dropped). The acceleration due to gravity (a) is approximately 9.8 m/s^2 downwards (assuming no air resistance). The displacement (s) is the initial elevation of the bomb, given as 310 m.
Plugging in the values, the equation becomes:
310 = 0 * t + 0.5 * 9.8 * t^2
Simplifying the equation, we have:
4.9 * t^2 = 310
Solving for t, we get:
t^2 = 310 / 4.9
t^2 = 63.265
Taking the square root of both sides, we find:
t ≈ 7.96 seconds
Therefore, the time required for the bomb to reach the earth is approximately 7.96 seconds.
b) How far does it travel horizontally while falling?
Since the helicopter is flying horizontally and the bomb is dropped while moving, the horizontal component of the bomb's velocity is the same as the helicopter's velocity. So we need to find the horizontal distance traveled by the bomb in time t.
The horizontal distance (d) traveled is given by:
d = v * t
where v is the horizontal velocity and t is the time.
In this case, the horizontal velocity (v) is given as 75.0 m/s.
Plugging in the values, we have:
d = 75.0 * 7.96
Therefore, the bomb travels horizontally approximately 597 meters while falling.
c) Find the horizontal component of the bomb's velocity just before it strikes the earth.
As mentioned earlier, the horizontal component of the bomb's velocity is the same as the helicopter's velocity, which is given as 75.0 m/s.
Therefore, the horizontal component of the bomb's velocity just before it strikes the earth is 75.0 m/s.
d) Find the vertical component of the bomb's velocity just before it strikes the earth.
To find the vertical component of the bomb's velocity just before it strikes the earth, we can use the equation of motion for vertical motion:
v = u + a * t
where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
In this case, we know the initial vertical velocity (u) is 0 m/s and the acceleration due to gravity (a) is approximately 9.8 m/s^2 downwards. We have already calculated the time (t) as approximately 7.96 seconds.
Plugging in the values, the equation becomes:
v = 0 + 9.8 * 7.96
Therefore, the vertical component of the bomb's velocity just before it strikes the earth is approximately 77.808 m/s downwards.
e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?
Since the velocity of the helicopter remains constant, it will continue to travel horizontally at a speed of 75.0 m/s during the time the bomb is falling.
Using the formula for distance, we can calculate the horizontal distance (d) traveled by the helicopter during the time it takes the bomb to reach the ground:
d = v * t
where v is the horizontal velocity of the helicopter and t is the time it takes for the bomb to reach the ground.
Plugging in the values, we have:
d = 75.0 * 7.96
Therefore, the helicopter will be approximately 597 meters away horizontally when the bomb hits the ground.