A projectile is launched with an initial speed of Vo at an angle θ above the horizontal. It lands at the same level from which it was launched. What was the average velocity between launch and landing? Explain please because I can't understand this.

1 Answer
Apr 25, 2016

Answer:

#"Average velocity"=V_@ cos theta#

Explanation:

enter image source here
Typical flight of a projectile is as shown in the picture above.
In the problem it is given that initial velocity #V_@# at an angle #theta# above the horizontal. As such inn the picture #"U"=V_@#.

This velocity can be resolved into its #x and y# components.
Component along #x# axis, and
Component along #y# axis#=V_@ sin theta#

We also know that both #x and y# components are orthogonal or perpendicular to each other, therefore can be treated separately.

Maximum height is achieved due to #sin theta# component of the velocity and Horizontal range is achieved due to #cos theta# component.

#sin theta# component.
This component of the velocity decreases due to action of gravity. Becomes zero at the maximum height point. Then increases due to gravity and becomes equal to initial #sintheta# component but in the opposite direction. We have ignored the friction due to air (Drag) acting on the projectile.
Let #t# be time of flight.
#"Average velocity"="Displacement"/"Time of flight"#

It is given that "It lands at the same level from which it was launched", means that displacement in the #y# axis is #=0#. From above equation we obtain
#"Average velocity"=0/t=0# .....(1)

#cos theta# component.
If we ignore air resistance, #cos theta# component of velocity #=V_@ cos theta# remains constant throughout the time of flight. Therefore,
#"Average velocity"=V_@ cos theta# .....(2)

Now to find the Resultant Average velocity we need to add both vectors along #x and y# direction. In this instant it is simple as one of the vectors is #=0#.

Hence, #"Average velocity"=V_@ cos theta#