A plane 65 m above the ground is flying directly toward a target at 50 m/s .
Part A
At what distance from the target should the pilot drop the weight?
h = 0.5g*t^2.
65 = 4.9t^2, t = ?.
d = Vo*t = 50 * t =
To find the distance from the target where the pilot should drop the weight, we need to consider the time it takes for the weight to fall. Since the plane is flying directly toward the target, we can assume the horizontal distance traveled by the plane is the same as the distance from the target.
Before we calculate the time, it's important to note that the motion of the weight falling can be divided into horizontal and vertical components.
Since the weight is dropped vertically with no initial horizontal velocity, the horizontal component of the motion remains unchanged at 50 m/s.
The vertical motion of the weight can be calculated using the equation of motion:
y = y0 + v0*t + (1/2)*a*t^2
Where:
- y is the vertical position of the weight
- y0 is the initial vertical position (65 m above the ground)
- v0 is the initial vertical velocity (0 m/s since it's dropped)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time passed since the weight is dropped
Since the weight is dropped from rest, the equation simplifies to:
y = y0 + (1/2)*a*t^2
Substituting the known values, we have:
y = 65 + (1/2)*(-9.8)*t^2
The weight hits the ground when y becomes zero, so:
0 = 65 + (1/2)*(-9.8)*t^2
Rearranging the equation, we get:
(1/2)*(-9.8)*t^2 = -65
Simplifying further, we have:
t^2 = (-65)*2/(-9.8)
t^2 = 13.26
Taking the square root of both sides, we find:
t ≈ √13.26
t ≈ 3.64 s (rounded to two decimal places)
Since the horizontal distance covered by the plane is equal to the distance from the target, we can determine the distance by multiplying the horizontal speed of the plane by the time:
Distance from target = (50 m/s) * (3.64 s)
Distance from target ≈ 182 m (rounded to the nearest whole number)
Therefore, the pilot should drop the weight approximately 182 meters from the target.