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5 days ago

7

A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At

15.4 s after blastoff, the engines fail completely so the only force on the rocket from then on is the pull of gravity.
A) What is the max height the rocket will reach above the launch pad?
B) How fast is the rocket moving at the instant before the it crashes onto the launch pad?
C) How long after engine failure does it take for the rocket to crash onto the launch pad?

1 answer:

kicyunya [1K]5 days ago

7 0

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + $\frac{1}{2}$ a $t^{2}$

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + $\frac{1}{2}$ (2.25 x $15.4^{2}$ )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s² (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

$v^{2}$ = $u^{2}$ + 2gs

$0^{2}$ = $34.65^{2}$ + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m + 61.19 m = 328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying $v^{2}$ = $u^{2}$ + 2gs

$v^{2}$ = $0^{2}$ + 2(9.81 x 328)

$v^{2}$ = 0 + 6435.36

v = $\sqrt{6435.36}$ = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

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