Astone 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.)

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

        ² = ² - 2 g y

         0 = ² - 2 g y

         y =² / 2g

          y = 12.73² / (2 9.8)

          y = 8.27 m

i guess it’s a and b

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.