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Projectile Motion of a golf ball off a cliff

I miseed a day of school when they taught this topic here is a sample question, If someone can help me learn how to do these types of questions I would be most grateful

Sam Dupher hits a gold ball horizontally off a 24.0m high cliff with a speed of 40.0 m/sec. How far from the base of the cliff will the ball strike the ground?

So far What I have is...

In Y Dir
Displacement = -24.0
acceleration = -9.81 m/s^2
Vi= 40 m/s
t=?

dx= 40m/s*t

See my problem is I do not know how to solve for t.
I would really like to understand this so if anyone can help ! please do
 
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1
TaurenOfBlight said:
I miseed a day of school when they taught this topic here is a sample question, If someone can help me learn how to do these types of questions I would be most grateful

Sam Dupher hits a gold ball horizontally off a 24.0m high cliff with a speed of 40.0 m/sec. How far from the base of the cliff will the ball strike the ground?

So far What I have is...

In Y Dir
Displacement = -24.0
acceleration = -9.81 m/s^2
Vi= 40 m/s
t=?

dx= 40m/s*t

See my problem is I do not know how to solve for t.
I would really like to understand this so if anyone can help ! please do
in Y dir the initial velocity is 0
in X dir initial velocity is 40 m/s

To get the amount of time the ball is in the air you just need to determine how long it takes the ball to travel 24 meters in the Y dir, you probably know the equation that gives the total displacement:
displacement = v_initial*t + 0.5*acceleration*t^2
so in Y dir:
24 = 0*t + 0.5*9.81*t^2
24 = 0.5*9.81*t^2
t = Square_root(48/9.81)

now you have the t that you can use in:
dx= 40m/s*t
 
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few mathmatical equation

this apply tom most projection motion from ground to ground

in verticle component y:

accelration: -g g equal 9.8m/s2
velocity: -gt+usin(d) { g as above t=time u=intial velicty d=degree to the horizontal of projection}

displacement: -0.5gt^2+utsin(d)+0

note if it's project at a buildng or a cliff the equation of displacemnt would be:
-0.5gt^2+utsin(d)+c where c is the meters above ground

in horizontal compent x:
acclearion =0 and always thats why horizontal motion is not affect by vertical compontent

velocity= ucos(d)
dispaclemtn = utcos(d)


fews thing needs to know :

the max displacemnt can be project with same intial velocity is when the degree of projection equals 45 degree

the max height (from ground level) of projection occur when time is equals to half
and vertical velocty equals to 0 plus horizontal displacemnt is half as well

horizontal velcity always is a constant

also if (degree of projection )is:
0 then sin(d)=0 cos(d)=1
90 then sin(d)=1 cos(d)=0
-90 then sin(d)=-1 cos(d)=0

ic in ur case the projection angle is -90
 
Last edited by a moderator:

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TaurenOfBlight said:
So far What I have is...

In Y Dir
Displacement = -24.0
acceleration = -9.81 m/s^2
Vi= 40 m/s
t=?
What's the initial speed in the y direction? It's not 40 m/s!

What's the equation for the vertical displacement as a function of time? (The vertical motion is uniformly accelerated (a = -g).)
 
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241
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btw projection motion are true upsidedown parabola
 

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