a plane at an altitude of 625 m is traveling at 95 m/s and drops a care package in inhabitants of a remote island in the south pacific.
a) how long will it take the package to hit the ground?
b)how far before the target should the pilot release the package?
c) what velocity does the package hit the ground (vector sum)
For the first one use h=ut;625=95*t;t=625/95
t=6.5789s
thank, but i figured that out.
what about the second one?
part b- vx=dx/t, manipulate the equation to find the horizontal velocity-dx=(vx)(t), then substitute the equation. dx=(95m/s)(6.6)=627m
To answer these questions, we need to consider the motion of the package in free fall. We can use the kinematic equations to solve each part:
a) How long will it take the package to hit the ground?
The first step is to find the time it takes for the package to reach the ground. We can use the equation:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s (since the package is dropped), the acceleration is the acceleration due to gravity (approximately 9.8 m/s^2), and the displacement is the altitude of the plane (625 m). Solving for t, we have:
625 = 0*t + (1/2)*9.8*t^2
Rearranging the equation, we get:
4.9*t^2 = 625
Dividing both sides by 4.9, we have:
t^2 = 625/4.9
Taking the square root of both sides, we find:
t ≈ √(625/4.9) ≈ 12.73 seconds
Therefore, it will take approximately 12.73 seconds for the package to hit the ground.
b) How far before the target should the pilot release the package?
To answer this question, we need to consider the horizontal motion of the package. Since there is no horizontal force acting on the package, its horizontal velocity remains constant throughout the flight. The horizontal distance traveled by the package is given by:
d = vt
where d is the horizontal distance, v is the horizontal velocity, and t is the time it takes for the package to reach the ground (which we found in part a). The horizontal velocity is equal to the velocity of the plane, which is given as 95 m/s. Substituting the values into the equation, we have:
d = 95 * 12.73 ≈ 1210.35 meters
Therefore, the pilot should release the package approximately 1210.35 meters before the target.
c) What velocity does the package hit the ground (vector sum)?
When the package hits the ground, it will have both vertical and horizontal components of velocity. The vertical velocity can be found using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s (since the package is dropped), the acceleration is the acceleration due to gravity (approximately 9.8 m/s^2), and the time is the value we found in part a (t ≈ 12.73 seconds). Solving for v, we have:
v = 0 + 9.8 * 12.73 ≈ 124.75 m/s
Therefore, the package hits the ground with a vertical velocity of approximately 124.75 m/s. To find the magnitude of the total velocity (the vector sum), we need to use the Pythagorean theorem:
total velocity = √(horizontal velocity^2 + vertical velocity^2)
Using the values we have, we can calculate:
total velocity = √(95^2 + 124.75^2) ≈ 157.37 m/s
Therefore, the package hits the ground with a total velocity (vector sum) of approximately 157.37 m/s.