Answer to Question #226624 in Mechanical Engineering for Papi Chulo

Question

A jet fighter aircraft is climbing vertically ay 120m/s when it releases a bomb that hits the ground 35seconds later. How high was the jet when the bomb was released?

Answer 1

Assume that the jet was at a height of h. Now since the jet is ascending vertically with a speed of 120m/s

Thus the bomb dropped will have the same speed initially and in the direction vertically upward. Assume the point of dropping to be the origin.

$So, S = ut +\frac{ gt^2}{2}\\ => h = 120m/s * 35s - 9.8m/s^2 *\frac{ 352}{2}\\ => h = -1802.5m$

This means that the ground is 1802.5m down the point of dropping.

Thus the height of jet when bomb dropped is 1802.5m.