Question

An airplane is in level flight at a speed of 400 km/h and an altitude of 1000 m when it drops a bomb. (a) find the time it takes the bomb to reach the bottom . (b) find the bomb's speed when it strikes the ground.

Answer 1

First, convert the speed from km h^-1 to m s^-1.\[\frac{400}{3.6} \approx 111.11 m s^{-1}\]When the plane drops the bomb, the bomb experiences a speed of 111.11 m s^-1 horizontally due to inertia. Using the formula, \[s=ut+\frac{at^{2}}{2}\]The horizontal displacement of the bomb will be\[s _{x}=111.11t\]Taking the upwards direction as positive, the vertical displacement of the bomb will be\[s_{y}=-\frac{g t^{2}}{2}\]For the bomb to reach the bottom, it needs to travel a vertical displacement of -1000 m.\[-1000=-\frac{g t^{2}}{2}\]\[t=\sqrt{\frac{2000}{g}}\]Taking g to be 9.81 m s^-2,\[t=\sqrt{\frac{2000}{9.81}}\]\[t \approx14.3 s\]So, the time taken for the bomb to reach the bottom is 14.3 s. Next, using this formula,\[v=u+at\]The horizontal velocity of the bomb will be\[v_{x}=111.11\]The vertical velocity of the bomb will be\[v_{y}=-g t\]To find the speed, you need to use the pythagoras' theorem,\[|v|=\sqrt{v_{x}^{2}+v_{y}^{2}}\]\[|v|=\sqrt{111.11^{2}+(-g t)^{2}}\]\[|v|=\sqrt{111.11^{2}+(-9.81(14.3))^{2}}\]\[|v|\approx179ms^{-1}\]So, the speed of the bomb when it strikes the ground is 179 m s^-1.