Answer to Question #283672 in Mechanics | Relativity for BxJ

Question

A bomb is dropped from a plane which is flying with a constant horizontal velocity. Locate the bomb after 1 second in relation to the plane.

Answer 1

It’s directly below the plane.

Assuming that no speed has been lost by air drag.

Drag loss=0

Time(t)=1sec

it will continue at the same forward speed as the plane, but dropping lower with time.

We know that

Newton motion second law

$S=ut+\frac{1}{2}at^2\rightarrow(1)$

$u=0m/sec\\t=1sec\\g=9.8m/sec^2$

Equation (1) put value

$y=0+\frac{1}{2}\times9.8\times1^2\\y=\frac{1}{2}\times9.8\times1^2m\\y=4.9m\\$