Question

You are outside on a windy day and you throw a 500g ball up in the air. You release the ball at a height of 1.2m above the ground. When you release the ball it is moving vertically straight up with a speed of 10m/s. The wind applies a constant force on the ball to the right with a magnitude of 2N. Ignore any effect the air has on the vertical motion of the ball. Where does the ball land? Make sure you draw a diagram.

Please show detail in your work.

Answer 1

let
m = 500 g = 0.5 kg
h = 1.2 m
voy = 10 m/s
Fx = 2 N

let t is the time taken for the ball to reach the ground.

use, -h = voy*t + (1/2)*(-g)*t^2

-1.2 = 10*t + (1/2)*(-9.8)*t^2

on solving the above equation

we get, t = 2.154 s

acceleration of the ball in x-direction,
ax = Fx/m
= 2/0.5

= 4 m/s^2

displacement of the ball in x-direction,

x = vox*t + (1/2)*ax*t^2

= 0 + (1/2)*4*2.154^2

= 9.28 m <<<<<<<<<<<<----------------Answer