Question

an aeroplane is flying vertically upwards with a uniform speed of 500m/s. when it is at a height of 1000m above the ground a shot is fired at it with a speed of 700m/s from a point directly below it. what should be the uniform acceleration of the aeroplane now so that it may escape from being hit

Answer 1

Answer on Question #44293-Physics-Mechanics-Kinematics-Dynamics

An airplane is flying vertically upwards with a uniform speed of $v_{10} = 500\frac{m}{s}$, when it is at a height of $h_{10} = 1000m$ above the ground a shot is fired at it with a speed of $v_{2} = 700\frac{m}{s}$ from a point directly below it. What should be the uniform acceleration of the airplane now so that it may escape from being hit?

Solution

For escape from a shot when a shot and the airplane would have the same height then they have the same velocity. That's why the velocity of a shot relatively the airplane is zero.

$$h_{1} = h_{10} + v_{10}t + \frac{a t^{2}}{2}.$$ $$h_{2} = v_{2}t.$$ $$v_{2} = v_{1} = v_{10} + a t \rightarrow t = \frac{v_{2} - v_{10}}{a}.$$ $$h_{1} = h_{2} \rightarrow h_{10} + v_{10}t + \frac{a t^{2}}{2} = v_{2}t \rightarrow h_{10} - (v_{2} - v_{10})t + \frac{a t^{2}}{2} = 0.$$

Put $t = \frac{v_2 - v_{10}}{a}$

$$h_{10} - (v_{2} - v_{10})\left(\frac{v_{2} - v_{10}}{a}\right) + \frac{a\left(\frac{v_{2} - v_{10}}{a}\right)^{2}}{2} = 0 \rightarrow h_{10} - \frac{1}{2a}(v_{2} - v_{10})^{2} = 0$$ $$a = \frac{(v_{2} - v_{10})^{2}}{2h_{10}} = \frac{\left(700\frac{m}{s} - 500\frac{m}{s}\right)^{2}}{2 \cdot 1000m} = 20\frac{m}{s^{2}}.$$

Answer: $20\frac{m}{s^{2}}$

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