Accelleration and freefall question ! please help!I'm stuck?

Question

A test rocket is fired vertically upward from a well. A catapult gives it initial speed 80.0m/s at ground level. its engines then fire and it accelerates upward at 4.00m/s^2 until it reached the attitude of 1000m. At that point its engines fail and the rocker goes into free-fall. with the acceleration of -9.80m/s^2

a) how long is the rocket in motion above the ground
b) what is its maximum altitude?
what is its velocity just before it collides with the earth?

for all this you need to consider the motion while the engine is operating separately from the free fall motion

Answer 1

What is alluded in free fall is motion without ari resistance toward a gravitational mass.
Actually the fall does not come really free. There has to be a cause.
When a sky diver falls its the result of some one pushed him offf the plane. So push is the cause of the fall.

The reason he is falling is the push of gravity that pushes him down.

In the case of a rocket moving upward till its power is cut it than reached an altitude where the gravity power is greater that the inertial power of the rocket,it would not cause it to fall back toward the earth.

As the rocket falls its acceleration increases,it does not ramain constant.
As the acceleration increase the speed of the fall increase.Till it reaches its maximun at the surface of the earth as it collides with it. The acceleration of the rocket would then be = 9.81+ meter/seconds^2.
It is calculated by the formula :
acc = GMe/ (Re +h)^2
Where G is newton constatn
Me=mass of the earth
Re=diameter of the earth
h=altitude above the earth
The acceleration can then be calculated as h aproaches zero.

Answer 2

v0 = initial velocity = 80 m/s
a = acceleration = 4 m/s^2
T = time
s = 1000 m (the altitude when the motor cuts off)

Distance traveled s = v0T + (1/2)aT^2

1000 = 80T + 2T^2 and 2T^2 + 80T - 1000
T = (-80 +/- SQRT(6400 + 8000))/4
T = ( -80 +/- 120)/4 = 10 seconds (ignore the negative value)

Now the rocket coasts until it reaches max altitude.
v (at burnout) = v0 + aT = 80 + 4(10) = 120
Now gravity is acting downwards with an acceleration of -9.8 m/s^2 (g).
v = v0 - at = 120 - 9.8T = 0 at max altitude
T = 12.25 seconds

The maximum altitude reached is:
1000 + (1/2)gT^2 = 1000 + 4.9(12.25)^2
Altitude = 1735.31 meters

The rocket now falls to the earth:
s =(1/2)gT^2 = 1735.31
T = SQRT( 1735.31/4.9) = 18.82 seconds

time of powered flight = 10 seconds
time of coasting = 12.25 seconds
time to fall back to the ground = 18.82 seconds

Total time in motion above the ground = 41.07 seconds
Maximum Altitude = 1735.31 meters

Velocity at impace = at = 9.8*18.82 = 184.42 m/s

Answer 3

Break the voyage into two phases:

Phase 1: the time during which the engines are on;
Phase 2: the time after engine cutoff.

Use this formula to find the time "t1" at engine cutoff:

d1 = Vo(t1) + a(t1)²/2
(Hint: d1=1000m; Vo=80m/s; a=4m/s²)
(Hint#2: Use the quadratic formula, and ignore the negative solution.)

Use this formula to find the speed V1 at engine cutoff:

V1 = Vo + a(t1)

Use this formula to find the additional distance d2 that the rocket rises after engine cutoff:

d2 = (V1)² / (2g) [Notice we use "g" instead of "a" this time, because the rocket is now accelerating under the unfluence of gravity.]

Maximum height = d1 + d2 (that's the answer to (b))

To get the time "t2" of Phase 2, use this formula: (we use "–d1" because we're calculating the time it takes to get to d1 meters _below_ its initial altitude; i.e., the time it takes to get to the ground from altitude d1).

–d1 = V1(t2) + g(t2)²/2 [remember that g is negative]
(Again, ignore the negative solution of t2.)

The total time in the air is then t1+t2. That's the answer to (a)

Its velocity at collision (V3) can be determined by knowing its maximum height (d1+d2). Use this formula:

V3 = sqrt(2g(d1+d2))

.

Answer 4

The velocity after the end of 1000 m
V^2 = U^2 + 2as = 6400 + 2x4 x1000
V = 120m/s.
Time to travel 1000m = t1 = [v-u] / a = 4/ [120 -80] = 10s.
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The velocity after falling down
W^2 = V^2 ─ 2gh = 120^2 - 2x 9.8x [─1000]
W =184.39m/s vertically down. (Answer for b)
------------------------------- ------------------------------
Time to travel [─1000] m, t2 = [W- V] / 9.8
= {184.39 - [-120]} /9.8= 31 s
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Answer for a) = 31+10 = 41 s
Answer for b) = 184.39m/s vertically down
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Answer 5

You have already given the altitude, it is 1000m, it will hit the ground at a velocity of 140 mps