projectile motion

[braddy]

Hi,

Please somebody help me with this problem:

A boy wants to knock down a coconut with a rock and his slingshot. He observes that the coconut is about 3.0m above his slingshot and the tree is 4.0m away along the ground.
He knows from experience that the release speed of his rock is 20m/s.
How far above the coconut should he aim?
Please can someone help me?

thank you

B.

[haditya]

the equation of the projectile is given by
y = x.v.tan p - 0.5*g.x.x.(sec p)^2

put x=3,y=4 solve for p

the height above which to aim is given by 3*tanp

the 1st projectile equation is got by eliminating time t in the 2 equations given below
x=v*cos p * t
y=v*sin p * t -0.5g*t*t

[braddy]

Originally posted by haditya@Sep 17 2005, 09:16 AM
the equation of the projectile is given by
y = x.v.tan p - 0.5*g.x.x.(sec p)^2

put x=3,y=4 solve for p
Quoted post


Thankx I will try this