A package is dropped from a passing plane(moving in the horizantal direction) and produces a projectile motion

Question

A package is dropped from a passing plane(moving in the horizantal direction) and produces a projectile motion as it is falling to the ground. If it takes the package only 39.36 seconds until it hits the ground, how far above the ground was the plane when the package was released?

Answer 1

remember a formula, that when a projectile is thrown horizontally, then its time of flight, T = sq.root (2H / g), where H is the height from where it is projected, and g is the gravity.
as in your case, a projectile is being let go, or thrown horizontally ( it means that it will have the same horizontal speed as that of the plane ), this formula can be applied.
T = sq.root(2H / g)
=> 39.36*39.36 = 2H / 9.8
=> H = 7591.127 m
i assure u, that this is the correct answer.

THINGS TO NOTE-
1.even if the plane was moving not perfectly horizontally, the horizontal speed of package would be equal to the horizontal speed of plane.
2. if u were asked in this question, also to calculate that "what horizontal distance will the package cover, from the point of projection?" ....then u would have multiplied it with the horizontal speed of plane (u would be given that)
3. this is important-
no matter how the plane flies, the package will always be vertically below the plane during its flight. That is, the horizontal distance covered by the plane is same as that covered by the package in those 39.36 sec.

Answer 2

Not sure I necessarily agree with the other answerer.

I'm going to assume that for purposes of this question that air resistance is zero and the object is released with zero vertical velocity.

I would think you would set up a quadratic equation for the motion, where t is time and y is the original height:

y = 16 t^2 or y - 16 t^2 = 0

Plug t into the equation and solve for y

y - 16 * 39.36 * 39.36 = 0
y - 16 * 1549.2096 = 0
y - 24787.3536 =0

So the object would have been released at about 24,787 feet and would have fallen to 0 in 39.36 seconds >without wind resistance< , which is a quite a bit more than 386 meters.

Answer 3

F=ma

The only downward force on the falling object is gravity, so:

ma = -mg
a = -g

Velocity if is the first time integral of acceleration, so:

v = -gt + C1

C1 = the initial velocity, so it is 0 as the object is 'dropped' rather than thrown downward.

v = -gt

Position is the time integral of velocity, so:

y = -1/2gt^2 + C2

Where C2 is the initial position, which we will rename y0. To find the hight above the ground, we make y=0 be the ground, and solve for y0 using our value of time t=39.36s, and using g=9.8m/s^2.

0 = -4.9t^2 + y0
y0 = 4.9(39.36)^2
y0 = 7591.12704 meters

Answer 4

If the piece separates gently (and not with an explosion as in case of a hearth cracker rocket ) then it is going to fly part by part as no exterior rigidity has acted on the stone or the piece. So the impressive answer is selection a million. In case of an explosion halfway it is going to matter near to separation. The small piece can bypass in any random course (no longer inevitably horizontal) and stick to diverse parabolic course.

Answer 5

Pardon my metric system use.

Gravity = 9.81 m/s^2

It accelerates according to gravity regardless of the projectile motion.

d = at
=(9.81)(39.36)
= 386.12 m

Answer 6

time of flight of the projectile is suppose t. the time of flight of the projectile as stated in your problem is given by t=sq.rt(h/g). u said time was 39.6 secs well then h/g = 39.6*39.6. thus, h= 39.6*39.6*9.8, thus h= 1515367.968 m