Question

A plane flying in a horizontal path 90m above the ground with a speed of 64 m/s², at what horizontal from the target should the pilot release the projectile?

Answer 1

Let's first find the time that the projectile takes to reach the ground:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot90\ m}{9.8\ \dfrac{m}{s^2}}}=4.28\ s.$$

Finally, we can find the horizontal distance from the target at which the pilot should release the projectile:

$$x=v_0t=64\ \dfrac{m}{s}\cdot4.28\ s=274\ m.$$

Answer:

$x=274\ m.$