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.

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Apr 25, 2016

Answer:

#"Average velocity"[email protected] 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 #[email protected]# at an angle #theta# above the horizontal. As such inn the picture #"U"[email protected]#.

This velocity can be resolved into its #x and y# components.
Component along #x# axis, and
Component along #y# axis#[email protected] 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 #[email protected] cos theta# remains constant throughout the time of flight. Therefore,
#"Average velocity"[email protected] 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"[email protected] cos theta#

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Write your answer here...
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Apr 15, 2016

Answer:

Average velocity = 0
Average speed = #V_o/2# where #V_o # = initial velocity

Explanation:

Average velocity = displacement# / time = 0 since the displacement is zero. The initial and final positions are the same.

Time to reach maximum height after launch = #V_o/g # where g = 9.8# m/(s^2)#
Time to land from maximum height is the same. Therefore, the total time from launch to land is #(2V_o)/g#

Maximum height = #(V_o^2)/(2g)#

Total distance traveled= launch to # h_max# and back = # (V_o)^2/g#

Average speed = total distance /total time
Average speed = #(V_o)^2/g# divided by #(2V_o)/g# which gives us #V_o/2#

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