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

Answer:

1.65 m

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:

$t=\frac{x}{v_x}=\frac{7.0 m}{12.0 m/s}=0.58 s$

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:

$S=\frac{1}{2}at^2=\frac{1}{2}(9.8 m/s^2)(0.58 s)^2=1.65 m$