Question

The position of a projectile fired with an initial velocity V0 feet per second and at angle Θ to the horizontal at the end of t seconds is given by the parametric equations x=(V0cosΘ)t, y=(V0sinΘ)t-16t^2. Suppose the initial velocity is 3 feet per second.
Obtain the rectangular equation of trajectory and identify the curve.

Answer 1

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 θ above the horizontal. As such inn the picture U=V∘.

This velocity can be resolved into its xandy components.

Component along x axis, and

Component along y axis=V∘sinθ

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

Maximum height is achieved due to sinθ component of the velocity and Horizontal range is achieved due to cosθ component.

sinθ 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 sinθ 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=DisplacementTime 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=0t=0 .....(1)

cosθ component.

If we ignore air resistance, cosθ component of velocity =V∘cosθ remains constant throughout the time of flight. Therefore,

Average velocity=V∘cosθ .....(2)

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

Hence, Average velocity=V∘cosθ

Hope it helps

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