impossible physics question? 10 pts to first correct answer.?

Question

A jet plane traveling horizontally at 150ms^-1 drops a rocket from a very tall height. The rocket immediately fires its engines after it is released from the plane, causing a 10.0ms^-2 (acceleration) in the X-DIRECTION (no x=vt), while falling under the influence of gravity in the y-direction. After the rocket has fallen 250 m, find:

a) the velocity in the y-direction
b) the velocity in the x-direction
c) the magnitute and direction of its velocity
d) how far it has traveled horizontally.

Neglect lift & drag forces.

THANKS!!! :)

(work is also appreciated)

i'm not trying to get out of doing my homework. i'm in 6 AP classes and sometimes i don't have time to ask the physics teacher for help. i've tried this problem for hours i just don't understand how to do it. and its not even homework...it was an example in my textbook.

Answer 1

Take downwards and forwards as positive and acceleration due to gravity = 9.8m/s^2
Rocket is travelling in same direction as plane (positive x direction)
Initial horizontal velocity = 150 m/s
Horizontal acceleration = 10 m/s^2
Initial vertical velocity = 0
Vertical acceleration = 9.8 m/s^2 (gravity)

a) v in y direction
Consider vertical velocities. There is only the acceleration due to free fall in the vertical direction
Initial vertical velocity = 0
After falling 250 m,
use v^2 = u^2 + 2as, but u = 0
v^2 = 2as
v = sqrt(2 x 9.81 x 250)
v = 70 m/s

b) v in x direction.
Time to fall 250 m: consider vertical components
Use v = u + at (but u = 0)
v = at
t = v/a
t = 70/9.8
t = 7.14 s
Consider horizontal velocities only.
Initial horizontal velocity is 150 m/s from plane
Use v = u + at
v = 150 + 10(7.14)
v = 221.4 m/s

c) Magnitude = sqrt(v(x)^2 + v(y)^2)
Magnitude = sqrt(221.4^2 + 70^2)
Magnitude = 232.2 m/s
Direction of velocity:
tan (angle) = 70/221.4
angle = arctan (70/221.4)
angle = 17.5 deg from horizontal

d) Time to fall 250 m: consider vertical components
Use v = u + at (but u = 0)
v = at
t = v/a
t = 70/9.8
t = 7.14 s
For horizontal range, consider horizontal components
s = ut + (1/2)at^2
s = 150(7.14) + (1/2)(10)(7.14^2)
s = 1326 m

10 points? =)

Answer 2

What's so impossible?

First, you have to find the time for the rocket to drop 250 m. If the rocket is acclerating horizontally, it has no effect on the accleration due to gravity. So the problem would be the same as an object being dropped 250 m from rest.
The time is 250= 4.9 m/sec^2 * t^2 and t is about 7.2 seconds.
The y direction velocity is then vy(t) = at, where t is the 7.2 sec from above.
The x direction velocity is then vx(t)=vo + a*t, where vo is the 150 m/sec speed of the jet and t is the 7.2 sec from above .
The velocity vector has a magnitude of
sqrt(vx(t)^2 + vy(t)^2) and a direction from the horizontal given by
tan-1[vy(t)/vx(t)] . Find the ratio and look up the angle in the tangent table for the ratio.
Horizontal distance is vo*t + 1/2 * 10*t^2

Answer 3

Accelerating is the cost at which the speed of an merchandise alterations. whilst a ball is tossed into the air gravity in the present day starts to act on the ball with the intention to advance up it. on account that gravity basically acts interior the downward direction, gravity can basically generate a metamorphosis interior the speed of a freefalling merchandise interior the downward direction! It would not surely count if the ball is shifting upward, downward or is at relax. however the ball is doing at t=0 seconds, one 2nd later the speed of the ball would be 9.8m/s extra downward than it replaced into on the beginning up. to illustrate, if a ball is shifting upward at 50 meters in line with 2nd whilst it fairly is released, after one million 2nd the speed of the ball gets replaced through 9.8 m/s. Vf = 50-9.8 = 40.2 m/sec. So after the only million 2nd the ball's upward speed gets replaced in teh downward direction through 9.8m/s. If the ball is released from relax, after one million 2nd the gravitational tension could have back replaced its speed downward through 9.8m/s. Vf = 0-9.8 -9.8 m/s! If the ball is movinf downward at -40 m/s on the 2nd it fairly is released, after one million 2nd that speed gets replaced downward through 9.8m/s. Vf = -40-9.9=-40 9.8 m/s. In all 3 situations the exchange interior the speed of the object replaced into interior the downward direction! evaluate what might ensue if this replaced into not authentic! assume for a 2nd that the gravitational acceleration replaced right into a superb 9.8m/s^2! in case you released a ball under those circumstances its speed might exchange interior the upward direction! launch a ball drom relax and after one million 2nd the ball might then be shifting UPWARD at 9.8 m/s! i visit wager which you will possibly locate that to be fairly atypical.