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The projectile motion is fired with velocity of magnitude vo at the angle feta. Find feta for which the maximum elevation of th...Asked by Whitney
The projectile motion is fired with velocity of magnitude vo at the angle feta. Find feta for which the maximum elevation of the projectile is twice its range
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Answered by
bobpursley
feta is a cheese, not much else. Theta is a Greek letter that is often used for an angle. But feta? it won't hold up more than about 30 degrees, and even then if falls down.
Vertical height=2*horiztal distance.
So vertical height first:
at the top, vertical velocity is zero.
vtop=vo*sinTheta-g t
or time to top is vo*sinTheta/g
then verticalmax height is
hmax=vo*sinTheta*vo*sinTheta/g
horizontal distance:
vo*cosTheta* vo sinTheta/g
then
vertical max=2*horizontal
(vo*sinTheta)^2 /g = 2*(vo^2 cosTheta*sinTheta)/g
or sintheta=2costheta
tan Theta=2
theta=arcTan2 or = 63.4349488 degrees
check all this.
Vertical height=2*horiztal distance.
So vertical height first:
at the top, vertical velocity is zero.
vtop=vo*sinTheta-g t
or time to top is vo*sinTheta/g
then verticalmax height is
hmax=vo*sinTheta*vo*sinTheta/g
horizontal distance:
vo*cosTheta* vo sinTheta/g
then
vertical max=2*horizontal
(vo*sinTheta)^2 /g = 2*(vo^2 cosTheta*sinTheta)/g
or sintheta=2costheta
tan Theta=2
theta=arcTan2 or = 63.4349488 degrees
check all this.
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