# 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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#### Explanation

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#### Explanation:

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Apr 25, 2016

"Average velocity"[email protected] cos theta

#### Explanation:

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 $x \mathmr{and} y$ components.
Component along $x$ axis, and
Component along $y$ axis[email protected] sin theta

We also know that both $x \mathmr{and} y$ 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 $\sin \theta$ 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.
$\text{Average velocity"="Displacement"/"Time of flight}$

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
$\text{Average velocity} = \frac{0}{t} = 0$ .....(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 $x \mathmr{and} y$ 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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#### Explanation

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#### Explanation:

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3
Apr 15, 2016

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$\frac{m}{{s}^{2}}$
Time to land from maximum height is the same. Therefore, the total time from launch to land is $\frac{2 {V}_{o}}{g}$

Maximum height = $\frac{{V}_{o}^{2}}{2 g}$

Total distance traveled= launch to ${h}_{\max}$ and back = ${\left({V}_{o}\right)}^{2} / g$

Average speed = total distance /total time
Average speed = ${\left({V}_{o}\right)}^{2} / g$ divided by $\frac{2 {V}_{o}}{g}$ which gives us ${V}_{o} / 2$

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