An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ 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 ($g = 10 ms^{-2}$)
An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ 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 ($g = 10 ms^{-2}$)
$1:5$
$5:1$
$1:40$
$40:1$
An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ 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 ($g = 10 ms^{-2}$)
Standard equation of projectile motion
$y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
Comparing with given equation
$A = \tan \theta $ and $B = \frac{g}{{2{u^2}{{\cos }^2}\theta }}$
So $\frac{A}{B} = \frac{{\tan \theta \times 2{u^2}{{\cos }^2}\theta }}{g} = 40$ (As $\theta = 45^\circ ,\;u = 20\,m/s,\;g = 10\,m/{s^2}$)
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