A box containing food supplies for a refugee camp was dropped from a helicopter flying horizontally at a constant elevation of 414.1 m. If the box hit the ground at a distance of 95.3 m horizontally from the point of its release, ...
a) what was the speed of the helicopter?
b) with what speed did the box hit the ground?
To solve this problem, we can use kinematic equations to find the necessary information.
a) To find the speed of the helicopter, we need to find the time it took for the box to hit the ground horizontally.
First, let's consider the horizontal motion. We know that the horizontal distance traveled by the box, x, is 95.3 m. We also know that the helicopter is flying at a constant horizontal speed. This means that the horizontal velocity of the box is the same as the horizontal velocity of the helicopter.
Using the equation for horizontal distance, we have:
x = v_h * t
Where:
x is the horizontal distance traveled by the box (95.3 m),
v_h is the horizontal velocity of the helicopter,
t is the time it took for the box to hit the ground.
Now, let's consider the vertical motion of the box. We know that the box was dropped from a height of 414.1 m.
Using the equation for vertical distance, we have:
h = 1/2 * g * t^2
Where:
h is the vertical distance traveled by the box (414.1 m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time it took for the box to hit the ground.
Rearranging this equation, we can solve for t:
t = sqrt(2h / g)
Now, we can substitute the value of h:
t = sqrt(2 * 414.1 / 9.8)
Calculating the value of t, we find:
t ≈ 9.02 s
Now, substitute the values of x and t into the first equation to find the horizontal velocity of the helicopter:
95.3 = v_h * 9.02
Simplifying:
v_h = 95.3 / 9.02
Calculating the value of v_h, we find:
v_h ≈ 10.57 m/s
Therefore, the speed of the helicopter is approximately 10.57 m/s.
b) To find the speed at which the box hit the ground, we can use the time, t, calculated earlier.
Using the equation for vertical velocity, we have:
v_v = g * t
Where:
v_v is the vertical velocity of the box when it hits the ground.
Substituting the values of g (9.8 m/s^2) and t (9.02 s), we find:
v_v = 9.8 * 9.02
Calculating the value of v_v, we find:
v_v ≈ 88.996 m/s
Therefore, the box hits the ground at a speed of approximately 88.996 m/s.
a. h = 0.5g*Tf^2.
h = 414.1 m.
g = 9.8 m/s^2.
Tf = ? = Fall Time.
Dx = Xo*Tf.
Dx = 95.3 m.
Xo = ? = Initial hor. speed.
b. Y = Yo + g*Tf.
Yo = 0.
Y = ? = Ver. component of total velocity.
Speed = Sqrt(Xo^2 + Y^2).