Question

A lacrosse ball leaves the sick horizontally at 12.0m/s. the catcher is 14.0 away. How much does the ball fall vertically travelling halfway to the catcher?

Answer 1

1.65 m

Step-by-step explanation:

The motion of the ball is a projectile motion, so it is a parabolic motion with two independent motions:

- on the x-axis, a uniform motion with constant speed
`v_x=12.0 m/s`

- on the y-axis, a uniformly accelerated motion with constant acceleration
`a=9.8 m/s^2` downward.

First of all, we need to find the time t at which the ball has travelled halfway to the chatcher, i.e. at a distance of
`x=14.0/2=7.0 m`. This can be found by using the relationship between distance, time and velocity along the horizontal direction:

`t=(x)/(v_x)=(7.0 m)/(12.0 m/s)=0.58 s`

So now we can move to the vertical motion, and we can calculate the distance covered vertically by the ball while falling for t=0.58 s, by using:

`S=(1)/(2)at^2=(1)/(2)(9.8 m/s^2)(0.58 s)^2=1.65 m`