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

Step-by-step 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 =
`(gt^(2))/(2)`,

or, t =
`\sqrt{(2h)/(g)}`

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

d = vt

=
`v \sqrt{(2h)/(g)}`

=
`73.0 m/s \sqrt{(2 * 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.