Question #224619

During a test, a rocket travel upwards at 75m/s. when it is 40m from the ground, the engine fails. Determine the maximum height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81m.s2 due to gravity. Neglect the effect of air resistance.

Expert's answer

h = 40 m

Acceleration a = 75 m/s^2

To = time before engine falls

Therefore

Hence velocity of engine is given by:

Vo = a x to =

So velocity of rocket is

V = Vo â€“ gt

If v = 0 it will be maximum height

Hence t = Vo/g

Or h max = ho + vot â€“

= ho +

= ho + ha/g

= ho(1 + a/g)

= 40 (1 + 75 / 9.8)

= **345.8 m**

According to law of conservation of energy

mgh max = where v is the speed of rocket before it hits the ground

v =

=

=

** v = 82.4 m/s**

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