A rescue plane is flying horizontally at a height of 132 m above the ground. The pilot spots a survivor and releases an emergency kit with a parachute. The kit descends at a constant vertical acceleration of -6.89 m/s2 and the initial plane horizontal speed is 68.9 m/s. Assuming air resistance and wind are negligible, how far away from the survivor will the emergency kit hit the ground?
a) how long does it take to fall 132m? If it takes t seconds,
b) the kit lands 68.9t meters away from where it was dropped. No idea how far away from the survivor that happened.
To find the distance from the survivor where the emergency kit hits the ground, we need to break down the motion into horizontal and vertical components.
First, let's consider the vertical motion. The vertical acceleration is given as -6.89 m/s², which means the kit is accelerating downwards. The initial vertical velocity is 0 since it is released from rest. We can use the equation of motion for vertical motion:
h = ut + (1/2)at²
where h is the height, u is the initial velocity, a is the acceleration, and t is the time. Since the kit starts from rest, the equation simplifies to:
h = (1/2)at²
Plugging in the values, we have:
132 m = (1/2)(-6.89 m/s²)t²
Solving for time, t:
t² = (2 * 132 m) / -6.89 m/s²
t² = -38.32
t ≈ √(-38.32) (Ignoring the negative solution since time can't be negative)
t is not a real number which means the kit will never hit the ground vertically.
Now let's consider the horizontal motion. The horizontal speed of the plane, 68.9 m/s, remains constant throughout the whole motion. Since there is no horizontal acceleration, the horizontal distance covered can be calculated using the formula:
d = ut
where d is the distance and u is the initial velocity. Plugging in the values:
d = 68.9 m/s * 0 s
d = 0 m
Therefore, the emergency kit will never hit the ground horizontally either. It will continue to fall vertically without traveling horizontally.