A bomber flies with a constant velocity of 50 m/s horizontally and wants to hit a target traveling 20 m/s (same direction) on a highway 1000 m below. What is the horizontal distance of the bomber from the target so that a bomb released from it will hit the target?

###### Solution.

The position-time equation of vertical motion of the bomb is given by

y_{f} = y_{i} – (g∙t^2)/2

We can calculate the time required for the bomb to hit the target. We substitute the values into the equation and solve for t

0=1000-(9.8∙t^2)/2

t≈14.29 s

The bomb is approaching the moving target, so we can use the relative velocity method. The method allows us to calculate the horizontal velocity of the bomb with respect to the target regarded as being at rest. The relative horizontal velocity of the bomb with respect to the target is given by

v_{BT} = v_{B} – v_{T}

where v_{BT} – the horizontal velocity of the bomb with respect to the target, v_{B} – the absolute horizontal velocity of the bomb and v_{T} – the absolute velocity of the target.

Hence,

v_{BT} = 50 – 20 = 30 m/s

This is how fast the bomb is approaching the target (horizontally).

In order to calculate the horizontal distance of the bomber from the target we use the position-time equation of horizontal motion of the bomb

x_{f} = x_{i} +v_{BT}∙t

Since, the horizontal distance is the horizontal displacement of the bomb we rewrite it as

∆x = x_{f} -x_{i} = v_{BT}∙t

Now we substitute the values into the equation and solve

∆x = 30∙14.29 = 428.7 m

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