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

R = 33.1 m and y = 8.27 m

Step-by-step 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