Question a6088

Nov 1, 2016

(a) $1376.27 m$, 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 ${v}_{\circ} = 97.5 m {s}^{-} 1$ 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 $\cos \theta \mathmr{and} \sin \theta$ 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

(a). To calculate time of flight $T$ when the package reaches the ground. Using the kinematic equation and taking $g = 9.81 m {s}^{-} 2$
$h = u + \frac{1}{2} g {t}^{2}$.
$y = \left({v}_{o} \sin \theta\right) T - \frac{1}{2} g {T}^{2}$
$\implies - 732 = \left(97.5 \times \sin 50\right) T - \frac{1}{2} \times 9.81 \times {T}^{2}$
$\implies - 732 = 74.383 T - 4.905 {T}^{2}$
$\implies 4.905 {T}^{2} - 74.383 T - 732 = 0$
This quadratic equation can be solved by using the formula for
$a {x}^{2} + b x + c = 0$, the roots are given by
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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 $T = 21.96 s$

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
$v = u + 2 a t$

Inserting given values we get
${v}_{g r o u n {d}_{y}} = \left({v}_{o} \sin \theta\right) + \left(- 9.81\right) T$
${v}_{g r o u n {d}_{y}} = \left(97.5 \times \sin 50\right) + \left(- 9.81\right) 21.96$
${v}_{g r o u n {d}_{y}} = - 141.045 m {s}^{-} 1$

Resultant of this calculated velocity and Horizontal component of velocity $\left({v}_{o} \cos \theta\right)$ gives the required angle.