Question

A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.

Answer 1

Answer:

(a) t = 4.52 sec

(b) X = 1,156.49 m

Explanation:

Horizontal Launching

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity $v_o$ remains the same in time. The distance x is computed as .

$\displaystyle x=v_o.t$

(a)

The vertical component of the velocity $v_y$ starts from zero and gradually starts to increase due to the acceleration of gravity as follows

$v_y=gt$

This means the vertical height is computed by

$\displaystyle h=h_o-\frac{gt^2}{2}$

Where $h_o$ is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

$\displaystyle t=\sqrt{\frac{2h_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec$

(b)

We now compute the horizontal distance knowing $v_o=256\ m/s$

$\displaystyle x=(256).(4.52)$

$X=1,156.49\ m$