Question

If the plane is flying in a horizontal path at an altitude of 98.0 m above the ground and with a speed of 73.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

Answer 1

Explanation:

The given data is as follows.

     height (h) = 98.0 m,     speed (v) = 73.0 m/s,

Formula of height in vertical direction is as follows.

          h = $\frac{gt^{2}}{2}$,

or,      t = $\sqrt{\frac{2h}{g}}$

Now, formula for the required distance (d) is as follows.

       d = vt

          = $v \sqrt{\frac{2h}{g}}$  

        = $73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}$  

          = 326.5 m

Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.