Question

A dive-bomber has a velocity of 280 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25 km. Find the angle θ.

Answer 1

Answer:

$\theta = 33.5 degree$

Explanation:

As we know that net displacement is

$d = 3.25 km$

altitude is given as

$h = 2.15 km$

so its horizontal displacement is given as

$x^2 + h^2 = d^2$

$x^2 + 2.15^2 = 3.25^2$

$x = 2.44 km$

now we have

$v_x = 280 cos\theta$

$v_y = 280 sin\theta$

now in x direction we have

$2.44 \times 10^3 = (280 cos\theta) t$

in y direction we have

$2.15 \times 10^3 = (280 sin\theta) t + 4.9 t^2$

from above equations we have

$2.15 \times 10^3 = 2.44 \times 10^3 tan\theta + 4.9 (\frac{2.44 \times 10^3}{280 cos\theta})^2$

by solving above equation

$\theta = 33.5 degree$