Answer to Question #224619 in Mechanical Engineering for Papi Chulo

Question

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.

Answer 1

h = 40 m

Acceleration a = 75 m/s^2

To = time before engine falls

Therefore $t0 = \sqrt(\frac{2h}{a})$

Hence velocity of engine is given by:

Vo = a x to = $\sqrt(2 h a)$

So velocity of rocket is

V = Vo – gt

If v = 0 it will be maximum height 

Hence t = Vo/g

Or h max = ho + vot – $\frac{gt^2}{2}$

       = ho + $\frac{V0^2}{g}$

      = ho + ha/g

       = ho(1 + a/g)

       = 40 (1 + 75 / 9.8)

       = 345.8 m

According to law of conservation of energy 

mgh max = $\frac{mv^2}{2}$ where v is the speed of rocket before it hits the ground

v = $\sqrt2 g h(1+\frac{a}{g})$

     = $\sqrt(2) h (a+g)$

     = $\sqrt(2 x 40 (75 + 9.81)$

v = 82.4 m/s