A catapult launches a test rocket vertically upward from a well, giving the rock

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 81.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.20 m/s2 until it reaches an altitude of 1060 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of ?9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.) (a) For what time interval is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it hits the ground?

Explanation / Answer

s=ut+(1/2)at^2

1060=81t + 0.5*4.2*t^2

=2.1t^2 + 81t -1060=0

t=10.32 sec

now engine fails

the velocity at this point is

v-u=at

v-81=4.2*10.32

v=124.34 m/s

now at top reaches after time t1

v-u=at

-124.34=-9.8*t1

t1=12.69 s

and distance travelled during this time t1

v^2 -u^2=2as

s=-124.34^2 /2*(-9.8)

s=788.8 m

thus total

height reached

S=1060+788.8=1848.8 m_____________(ans(b))

now tota time it takes to reach ground from here

s=ut+(1/2)at^2

1848.8 =0.5*9.8*(t2)^2

t2=19.42 sec

thus total time of flight

T=t+t1+t2=10.32 + 12.69 +19.42

=42.43 sec ________________(ans(a))

now just before hitting ground

v^2-u^2=2as

v^2=2*9.8*1848.8

v=190.36 m/s __________(ans(c))

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