Projectile Motion Kinematics Problem

[Turquoise]

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]

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.

 

[Lachlan1]

ok, so I am 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

 

[Lachlan1]

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 traveling 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.