An arrow shot horizontally from a cliff at 15 m/s lands 30 m away. How how high is the cliff ?
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OpenStudy (anonymous):
Projectile motion in 2D (x,y plane) : It is fairly lengthy to derive these equations:
OpenStudy (anonymous):
I cannot make an equation now...
OpenStudy (anonymous):
x=( v cos alpha)t
y=(v sin alpha)t -1/2 g t^2
Where alpha is the angle you shoot the arrow.
v is the initial velocity
t is time
g is the universal gravitational constant
OpenStudy (anonymous):
does it have to be so complex?
OpenStudy (anonymous):
The answer for this question is 19.6cm height i dont know how to solve .
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OpenStudy (anonymous):
you need the right formula i think
OpenStudy (anonymous):
There will be so many simplification as you are shooting the arrow horizontally.
Cos alpha=1
Sin alpha =0
OpenStudy (anonymous):
x=vt
y=-1/2 g t^2
OpenStudy (anonymous):
30=15 t
Thus t=2
OpenStudy (anonymous):
y=-1/2 g 2^2=-2g=-2*9.80=-19.6
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OpenStudy (anonymous):
I am correct ?
OpenStudy (anonymous):
Well you just had the solution :) And yes that is it
OpenStudy (anonymous):
well done andras
OpenStudy (anonymous):
Thanks for the help :)
OpenStudy (anonymous):
Somehow I could not create proper equations, so it does not look so neat
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