Prove that a guill shoot 3 times as high when it is fired at an angle of 60 degree as when it is fired at an angle of 30 degrees but at the same horizontal range?

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Answer 1 · 2018-06-06T17:12:11

See below:

The horizontal range is given by:

$sf(d=(v^2sin2theta)/g)$

When $sf(theta=30)$ then $sf(2theta=60)$

$sf(sin60=0.866)$

When $sf(theta=60)$ then $sf(2theta=120)$

$sf(sin120=0.866)$

This will give the same range for the 2 angles.

To find the height reached we can use:

$sf(v^2=u^2+2as)$

This becomes:

$sf(v^2=u^2-2gh)$

$:.$ $sf(0=(vsintheta)^2-2gh)$

$sf(h=(v^2sin^2theta)/(2g))$

When $sf(theta=60^@)$ :

$sf(h_(60)=(v^(2)0.75)/(2g))$

When $sf(theta=30^@)$ :

$:.$ $sf(h_(30)=(v^(2)0.25)/(2g))$

$:.$ $sf(h_(60)/h_(30)=(cancel(v^(2))0.75)/(cancel(2g))xx(cancel(2g))/(cancel(v^(2))0.25)=3)$