Question

Question #a6088

Answer 1

(a) `1376.27m`, rounded to two decimal places.

(b) Steps as below.

Explanation:

Imagine the following as shown in the figure below:
At the moment when the package is released it is traveling equal to the speed of the plane.
∴ Use the following.

Projectile (Package released by the pilot) has initial velocity of `v_@=97.5ms^-1` at an angle of `θ=50^@`with respect to the horizontal. `Δd_y=-732m`

Taking origin of coordinate system at the point of release of packet with positive directions as shown in the figure.

The package has velocity whose `x`-component and `y`-component can be found by multiplying initial velocity with `costhetaandsintheta` respectively.

Treat the component velocities of the package separately.
As we assume air friction is negligible there is no change of velocity in the `x`-direction.
Acceleration due to gravity is `−ve`
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(a). To calculate time of flight `T` when the package reaches the ground. Using the kinematic equation and taking `g=9.81ms^-2`
`h=u+1/2g t^2`.
`y=(v_osin theta) T-1/2 g T^2`
`=>-732=(97.5xxsin 50) T-1/2 xx9.81xx T^2`
`=>-732=74.383 T-4.905 T^2`
`=>4.905 T^2-74.383T-732=0`
This quadratic equation can be solved by using the formula for
`ax^2+bx+c=0`, the roots are given by
`x=(-b+-sqrt(b^2-4ac))/(2a)`

I calculated its roots using the inbuilt graphic utility. We plot this quadratic equation and find out value of `T` for `y=0`. Taking positive root only as time can not be negative. We get `T=21.96s`

`Δd_x="Horizontal component of velocity"xx"time of flight"`

`=>Δd_x=(97.5cos50)xx21.96`
`=>Δd_x=1376.27m`, rounded to two decimal places.

(b). To Calculate Vertical component of velocity as the package hits earth we use the kinematic equation
`v=u+2at`

Inserting given values we get
`v_(ground_y)=(v_osin theta)+(-9.81)T`
`v_(ground_y)=(97.5xxsin 50)+(-9.81)21.96`
`v_(ground_y)=-141.045ms^-1`

Resultant of this calculated velocity and Horizontal component of velocity `(v_o cos theta)` gives the required angle.