Answer to Question #120959 in Mechanics | Relativity for Zachery Goetz

Question #120959
A rocket rises vertically with acceleration of 3.2m/s^2. It runs out of fuel at 750 m. It’s acceleration from here is gravity.
1- what is the velocity of the rocket when it runs out of fuel.
2- how long does it take to reach 750 m.
3- what is the highest altitude the rocket reaches.
4- what velocity does it strike earth.
5- how long is it in the air
1
Expert's answer
2020-06-08T10:14:06-0400

Explanations & Calculations


  • The rocket starts from rest & after 750 m it enters a vertical motion under the gravity.
  • Therefore, until 750 m rocket is under 3.2 acceleration & just after 750 m its under the 9.8 ms-2 gravity.


1) Velocity after 750 m: apply V2 = U2 +2as vertically upward

V2=02+2×3.2ms2×750mV=4800m2s2V=403=69.28ms1\qquad\qquad \begin{aligned} \small V^2 & = \small 0^2 + 2\times 3.2ms^{-2} \times 750 m\\ \small V & = \small \sqrt{4800m^2s^{-2}}\\ V &= \small \bold{40\sqrt{3} =69.28ms^{-1}} \end{aligned}


2) Time taken to reach the 750 m : apply V = U +at

403ms1=0+3.2×tt=2532=21.65s\qquad\qquad \begin{aligned} \small 40\sqrt{3}ms^{-1} & = \small 0 +3.2\times t\\ \small t &= \small \bold{\frac{25\sqrt{3}}{2}=21.65s} \end{aligned}


3) Maximum height it reaches from 750 m : apply V = U +2as vertically upward (a = g =9.8 ms-2 )

02=(403ms1)2+2×(9.8ms2)×hh=4800m2s219.6ms2=244.898m\qquad\qquad \begin{aligned} \small 0^2 &= \small (40\sqrt3ms^{-1})^2 + 2\times(-9.8ms^{-2}) \times h\\ \small h &= \small \frac{4800m^2s^{-2}}{19.6ms^{-2}}\\ &= \small 244.898m \end{aligned}

Therefore, the highest altitude it reaches = 750 m +244.898 m = 994.898 m


4) Velocity, it strikes the ground with : apply V2 = U2 + 2as vertically downward from the 750 m height.

V2=(403)2+2×(+9.8)×(+750m)V=19500m2s2=10195=139.64ms1\qquad\qquad \begin{aligned} \small V^2 & = \small (-40\sqrt3 )^2 + 2\times (+9.8)\times(+750 m)\\ \small V &= \small \sqrt{19500 m^2s^{-2}}\\ &= \small \bold{10\sqrt{195}= 139.64 ms^{-1}} \end{aligned}


5) Time spent in free fall ,t : apply S = Ut +0.5st2 vertically downward from the height 750 m.

+750m=(403)×t+12×9.8×t20=9.8t2803t1500t=21.32s(or7.18s)\qquad\qquad \begin{aligned} \small +750 m &= \small (-40\sqrt3)\times t +\frac{1}{2} \times9.8 \times t^2\\ \small 0 &= \small 9.8t^2-80\sqrt3t-1500\\ \small t &= \small 21.32 s \,(\text{or} -7.18s )\\ \end{aligned}

Therefore, total time airborne T = 21.65s + 21.32s = 42.97s



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