A helicopter drops a box from 120m in the air.
a. Determine the time the box is in the air.
b. Determine the velocity the box would have right before it hits the ground.
c. Before the box falls the helicopter had a horizontal velocity of 55/s. Determine the distance the box will travel in the X-axis.
d. Solve for the resultant velocity that the box would impact the ground with.
I know it's a lot, but I'm really lost! Thanks in advance!(:
No problem! I'll help you with each part of the question step by step.
a. To determine the time the box is in the air, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = distance traveled (120m in this case)
u = initial velocity (0 m/s since the box was dropped)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time
Since the box was dropped, the initial velocity (u) is zero, and the equation simplifies to:
s = (1/2)at^2
Plugging in the values, we can solve for t:
120 = (1/2)(-9.8)t^2
Now, rearrange and solve for t:
t^2 = -(240 / -9.8)
t^2 = 24.5
t ≈ √24.5
t ≈ 4.95 seconds
So, the box is in the air for approximately 4.95 seconds.
b. To determine the velocity the box would have right before it hits the ground, we can use the kinematic equation again:
v = u + at
Where:
v = final velocity
u = initial velocity (0 m/s since the box was dropped)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time (4.95 seconds in this case)
Plugging in the values, we can solve for v:
v = 0 + (-9.8)(4.95)
v = -48.51 m/s
The negative sign indicates that the velocity is in the downward direction. So, the box would have a velocity of approximately -48.51 m/s right before it hits the ground.
c. Before the box falls, the horizontal velocity of the helicopter is given as 55 m/s. Assuming there is no horizontal acceleration, the distance (d) the box will travel in the x-axis can be calculated using the formula:
d = v * t
Where:
d = distance
v = velocity in the x-axis (55 m/s in this case, assuming it remains constant)
t = time (4.95 seconds)
Plugging in the values, we can solve for d:
d = 55 * 4.95
d ≈ 272.25 meters
So, the box will travel approximately 272.25 meters in the horizontal x-axis.
d. To solve for the resultant velocity that the box would impact the ground with, we can use the Pythagorean theorem:
resultant velocity = √(velocity in x-axis)^2 + (velocity in y-axis)^2
In this case, the velocity in the x-axis is 55 m/s (since there is no horizontal acceleration), and the velocity in the y-axis is -48.51 m/s (negative because it is in the downward direction).
Plugging in the values, we can solve for the resultant velocity:
resultant velocity = √(55)^2 + (-48.51)^2
resultant velocity ≈ √3025 + 2353.8401
resultant velocity ≈ √5378.8401
resultant velocity ≈ 73.34 m/s
So, the box would impact the ground with a resultant velocity of approximately 73.34 m/s.