A bomb is released from an airplane flying at a height H = 1…

A bomb is released from an airplane flying at a height $H = 1\ \mathrm{km}$ with speed $v_0 = 720\ \mathrm{km/h}$. To what time must its timer be set so that it explodes $h = 100\ \mathrm{m}$ above the ground?

Explanation of Method

Timer for a bomb to explode 100 m above the ground.

Step 1: Only the vertical motion matters

The bomb starts with zero vertical speed; its horizontal speed (200 m/s) does not affect how fast it falls. It must fall $H - h = 900\ \mathrm{m}$: \[H - h = \tfrac12 g t^2 \;\Longrightarrow\; t = \sqrt{\frac{2(H - h)}{g}} = \sqrt{\frac{1800}{9.81}}.\]

\[t \approx 13.55\ \mathrm{s}\]

$H = 1\ \mathrm{km},\ v_0 = 200\ \mathrm{m/s}$ $h = 100\ \mathrm{m}$ $t = 13.55\ \mathrm{s}$

Drawn to scale. The fall time depends only on the vertical drop (900 m), not on the plane's speed.

Result

\[\boxed{\,t \approx 13.55\ \mathrm{s}\,}\]

Remark

Note. By then the bomb has moved $200 \cdot 13.55 \approx 2.7\ \mathrm{km}$ forward (still directly below the plane) and would reach the ground only 0.73 s later. Compare with Gravitational Kinematics 034, where the fuse is set for a shell fired at an angle.