A project is projected with a velocity of $20\ m/s$ making an angle of $45{^\circ}$ with horizontal. The equation for the trajectory is $h = Ax - Bx^{2}$ where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A\ :B$ is $(g = 10\ ms^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(d) Standard equation of projectile motion
$y = x	an	heta - \frac{gx^{2}}{2u^{2}\cos^{2}	heta}$
Comparing with given equation
$A = 	an	heta$ and $B = \frac{g}{2u^{2}\cos^{2}	heta}$
So $\frac{A}{B} = \frac{	an	heta 	imes 2u^{2}\cos^{2}	heta}{g} = 40$
[As $	heta = 45{^\circ},\ u = 20\ m/s,\ g = 10m/s^{2}$]

Q. A project is projected with a velocity of $20\ m/s$ making an angle of $45{^\circ}$ with horizontal. The equation for the trajectory is $h = Ax - Bx^{2}$ where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A\ :B$ is $(g = 10\ ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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