# Projectile Motion Kinematics Problem

## Homework Statement

Will a ball, kicked at 14.0 m/s vertically and 9.0 m/s horizontally, clear a bar 3.0 m high and 20 m away from the kicker? Solve.

## Homework Equations

Kinematics Equations.

## The Attempt at a Solution

Haven't encountered this type of problem, didn't know where to start.

lightgrav
Homework Helper
Horizontal motion , and vertical motion, are independent ... linked by TIME .

re-word the question to make TIME explicit :
"will the ball be at least 3m high WHEN it gets to the bar at x=20m ?

After you figure out what time that happens, you can find its height.

ok, so im trying to learn these ones too. here is what i know. 14m/s vertical vector component of velocity. how high will the ball travel? well, this a kinematic equation problem.
Vo = initial velocity = 14m/s. (says in the question)
V = final velocity (when the ball is at the top of its flight path, this is) = 0
a = acceleration = -9.8m/s^-2 (only force acting on this object)
t = time = ?
x = change in x, or change in displacement, or change in distance vertically = ?

but with three knowns, we can find a fourth. in this case it would be good to see how high the ball is going to go, so we want to find 'change in x'

we know that
V^2 = Vo^2 + 2*a*x
this is on of the five kinematic equations derived from the ideas that velocity = change in distance/ change in time, and, acceleration = change in velocity/ change in time

so, rearanging mathematically this formula, to isolate x, we get
((V^2) - (Vo^2)) /2*a = x
now we can substitiute in the values we know to solve for x
((0-14^2))/2*-9.8 = x
10 = x is what i get. so this tells us the ball will get to a maximum height of 10m at some point. this is the first thing we need to know

so then, to find the time it takes for the ball to get to the maximum height
use the formula (V-Vo)/a = t
this gives a time of 1.43s

to find how far horizontally this point is from where the ball is kicked, now look at the horizontal component. 9m/s. the ball is travelling with a velocity of 9m for 1.43 s,
using the formula v = change in distance/change in time, rearange this to get
velocity*time = change in distance
9*1.43= 12.87m. so it has not yet reached the bar 20m away.