First slide

Projection Under uniform Acceleration

Question

A fighter plane enters inside the enemy territory at time t = 0, with velocity v₀ = 250 m/s and moves horizontally with constant acceleration a = 20 m/s² as shown in fig. (3).
An enemy tank at the border, spot the plane and fires shots at an angle θ = 60^o with the horizontal and with velocity u = 600 m/s. At what altitude H of the plane it can be hitted by the shot?

Difficult

A

$1500 \sqrt{3} m$
B

125 m
C

1400 m
D

2473 m

Solution

Let the fighter plane be hit at point Q.
Suppose its horizontal distance is d as shown in fig.
Now, $d = v_{0} t + \frac{1}{2} {at}^{2} = (ucos θ) t$
$∴ t = [\frac{ucos θ - v_{0}}{(a / 2)}] = \frac{600 \times (1 / 2) - 250}{(20 / 2)} = 5 s$
Further, $H = (usin θ) t - \frac{1}{2} {gt}^{2}$
$\begin{array}{l} = (600 \times \sqrt{3} / 2) \times 5 - \frac{1}{2} \times 10 \times 25 \\ = 1500 \sqrt{3} - 125 = 2473 m \end{array}$

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