Question #a4844

You must realise the movement on each axis. The suitcase will have an initial velocity equal to that of the airplane. This can be analysed on both axises:

#sin23^o=u_y/u#

#u_y=sin23^o*u=sin23^o*90=35.2m/s#

#cos23^o=u_x/u#

#u_x=cos23^o*u=cos23^o*90=82.8m/s#

Vertical axis

Note: You should aim towards finding the total time of motion on the vertical axis. Afterwards, the horizontal motion is easy.

The motion on the vertical axis is decelleration, since it initially goes up but gets pulled by gravity. After it reaches the maximum height, the motion is acceleration until it hits the ground. For the decelleration part, to find the time at which the maximum height is reached #t_1#

#u=u_(0y)-a*t_1#

Where:

initial speed is #u_y=35.2m/s#
acceleration is equal to #g=9.81m/s^2#
final speed is zero, since it changes direction at the peak #u=0#

#0=35.2-9.81*t_1#

#t_1=3.588# #s#

The height for decelleration is:

#h=h_0+u_0*t_1-1/2*a*t_1^2#

#h=114+35.2*3.588-1/2*9.81*3.588^2#

#h=177.15# #m#

Finally, the time for its free fall:

#h=1/2*g*t_2^2#

#t_2=sqrt((2h)/g)#

#t_2=sqrt((2*177.15)/9.81)#

#t_2=6# #s#

The total time:

#t_t=t_1+t_2#

#t_t=3.588+6#

#t_t=9.588# #s#

This is the total time it took for the suitcase to go upwards to a maximum height and then fall to the ground.

Horizontal axis

The speed on the horizontal axis is constant, since no forces are applied. For constant speed the distance on the horizontal axis as the object was falling (total time is common):

#s=u_x*t_t#

#s=82.8*9.588#

#s=793.89# #m#