Find out the answer to "A projectile is shot straight up from the..."

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Redemitz4s

Answered question

2021-11-08

A projectile is shot straight up from the earth’s surface at a speed of 10,000 km/h. How high does it go?

Answer & Explanation

George Blue

Beginner2021-11-09Added 18 answers

Given:
The speed of the projectile

v = 10000 k \frac{m}{h} = 2.78 \times 10^{3}

m/s
We need to determine the maximum height the projective can reach with given velocity.
For that we can apply conservation of energy. At maximum height, the object will have zero velocity and maximum potential energy.
Initial energy = Final energy

K_{i} + U_{i} = K_{f} + U_{f}

\frac{1}{2} m v_{i}^{2} - \frac{G M_{e} m}{R_{e}} = \frac{1}{2} m v_{f}^{2} - \frac{G M_{e} m}{R_{e} + h}

v_{f} = 0

- \frac{G M_{e} m}{R_{e} + h} = \frac{1}{2} m v_{i}^{2} - \frac{G M_{e} m}{R_{e}}

\frac{G M_{e}}{R_{e} + h} = - \frac{1}{2} v_{i}^{2} + \frac{G M_{e}}{R_{e}}

\frac{2 G M_{e}}{v_{i}^{2}} = \frac{R_{e}^{2}}{h} + R_{e}

\frac{R_{e}}{h} = \frac{2 G M_{e}}{v_{i}^{2}} - R_{e}

= \frac{2 \times 6.67 \times 10^{- 11} \times 5.98 \times 10^{24}}{7.72 \times 10^{6}} - 6.37 \times 10^{6}

= 103.3 \times 10^{6} - 6.37 \times 10^{6}

h = \frac{R_{e}^{2}}{96.63 \times 10^{6}}

= \frac{{(6.37 \times 10^{})}^{2}}{96.63 \times 10^{6}} = 4.2 \times 10^{5}

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