Question

A bomber plane is flying at a height “h” exactly level with the ground, and with a constant speed “v.” At time t = 0 the plane gently releases its payload (i.e. at the initial moment the payload is at rest with respect to the plane). Assume that after releasing the payload the plane continues to move with the same velocity as before. Answer the following questions based on this scenario. (You may use h = 490 meters and v = 70 meters/sec to obtain numerical answers.

a) how far along the ground does the bomb (payload) travel before it hits the ground?

b) Find the bombs velocity vector at the moment before it lands on the ground. You may choose to express it either as a magnitude and an angle.

c) At the instant that the bomb hits the ground what is its distance from the plane?

Answer 1

The horizontal velocity component of the payload remains constant = 70m/s

Time taken for the bomb to reach ground is

$t=\sqrt{\frac{2H}{g}}\\ \implies t=\sqrt{\frac{2\times 490}{9.8}}\\ \implies t=10s$

a) So the distance travelled is $R=vt\\ \implies R=70\times 10\\ \implies R=700m$

b) Final vertical component of velocity is

$v_y=u+gt\\ \implies v_y=0+10\times 10\\ \implies v_y=100m/s$

So final velocity vector is $(70\hat{i}-100\hat{j})m/s$

c) Distance is

$D=\sqrt{700^2+490^2}\\ \implies D=854.46m$