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- 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

Expert's answer

**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 V^{2} = U^{2} +2as vertically upward

"\\qquad\\qquad\n\\begin{aligned}\n\\small V^2 & = \\small 0^2 + 2\\times 3.2ms^{-2} \\times 750 m\\\\\n\\small V & = \\small \\sqrt{4800m^2s^{-2}}\\\\\nV &= \\small \\bold{40\\sqrt{3} =69.28ms^{-1}} \n\\end{aligned}"

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

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

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

"\\qquad\\qquad\n\\begin{aligned}\n\\small 0^2 &= \\small (40\\sqrt3ms^{-1})^2 + 2\\times(-9.8ms^{-2}) \\times h\\\\\n\\small h &= \\small \\frac{4800m^2s^{-2}}{19.6ms^{-2}}\\\\\n&= \\small 244.898m\n\\end{aligned}"

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

4) Velocity, it strikes the ground with : apply V^{2} = U^{2} + 2as vertically downward from the 750 m height.

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

5) Time spent in free fall ,t : apply S = Ut +0.5st^{2} vertically downward from the height 750 m.

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

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

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