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son4ous [18]
3 years ago
8

A stone is catapulted at time t = 0, with an initial velocity of magnitude 18.0 m/s and at an angle of 45.0° above the horizonta

l. (Neglect air resistance.) Find its horizontal and vertical displacements from the catapult site at the following times after launch. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)
Physics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

R = 33.1 m  and y = 8.27 m

Explanation:

This exercise is projectile launch, let's use the scope equation to find the displacement on the X axis

      R = vo2 sin 2θ / g

      R = 18 2 sin (2 45) /9.8

      R = 33.1 m

First let's look for the velocity components

       vox = vo cos θ

       voy = vo sin θ

       vox = 18 cos 45

       vox = 12.73 m / s

       voy = 18 sin 45

       voy = 12.73 m / s

To find the maximum vertical displacement, where the vertical velocity must be zero, we can use the equation

        v_{fy}² = v_{oy}² - 2 g y

         0 = v_{oy}² - 2 g y

         y =v_{oy}² / 2g

          y = 12.73² / (2 9.8)

          y = 8.27 m

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<u>Projectile Motion</u>

In a projectile motion (or 2D motion), the object is launched with an initial angle θ and an initial velocity vo.

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