Question #a6088

1 Answer
Nov 1, 2016

(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#
graphic tool

#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.