The airplane in the below figure is travelling at a horizontal speed of 79.32 m/s. Given the horizontal distance of to the mountain climbers of 425.0 m and the
airplane height of 235 m (above the mountain climbers), what must be the Initial
Vertical Component of the Velocity (m/s) so that the supply reaches the mountain
climbers?
Thrown upward?
0
235 m
Thrown downward
(< 0)
425 m

The projectile launch ratios allow finding the velocity for the package to reach the climbers, it must be thrown down with an initial velocity

          v_{oy}= 17.52 m / s

given parameters

to find

Projectile launching is the application of kinematics to motion in two dimensions where there is no acceleration on the x-axis. Let's set a reference frame where the x-axis is in the direction of the plane and the y-axis is vertical and positive upward.

Find the time it takes for the package to reach the climbers

                vₓ =

where vₓ is the horizontal velocity, x the displacement and t the time

               t =

               t = 425 / 79.32

               t = 5.358 s

This is the time for all movement since time is a scalar

now we look for the initial vertical velocity

             y = y₀ + v_{oy} t - ½ g t²

The package is at an initial height of y₀ = 235 m and when it reaches the climbers the height is zero (y = 0)

            0 = y₀ + v_{oy} t - ½ g t²

            v_{oy} = ½ g t - y₀/ t

            v_{oy} = ½ 9.8 5,358 - 235/ 5,368

            v_{oy} = 26.2542 - 43.778

            v_{oy} = -17.52 m / s

the negative sign indicates that the speed is down

For the package to reach the climbers, it must be thrown downward with an initial velocity of 17.52 m / s.

learn more about projectile launch here:

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a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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