Question

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 horizontal. (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.)

Answer 1

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