Projectile Motion Comparison - 2

Classical Mechanics Level pending

Two balls are thrown from the same point with the same speed, the first freely at an angle of $45^{\circ}$ with the horizontal, and the second is sent sliding on an inclined plane, with its initial velocity vector making an angle of $45^{\circ}$ with the horizontal edge of the plane. How do the maximum altitudes of the two balls compare ? The inclined plane makes an angle of $30^{\circ}$ with the horizontal ground. If $H_1$ is the maximum altitude of the first ball, and $H_2$ is the maximum altitude of the second ball, what is the relation between $H_1$ and $H_2$ ?

The attached figure depicts the two projectiles. The figure is not drawn to scale.

H_2 = 2 H_1

H_2 = H_1

H_2 = \frac{1}{2} H_1

1 solution

A Former Brilliant Member
May 19, 2020

First case :

$H_1=\dfrac{u^2\sin^2 45\degree}{2g}=\dfrac{u^2}{4g}$ .

Second case :

Acceleration $=g\sin 30\degree=\dfrac{g}{2}$ along the line of greatest slope of the inclined plane. Component of initial velocity along this line is

$u\sin 45\degree=\dfrac{u}{\sqrt 2 }$ ,

and along the horizontal line is

$u\cos 45\degree=\dfrac{u}{\sqrt 2 }$

So, $H_2=\dfrac{u^2}{2\times 2\frac{g}{2}}\times \sin 30\degree=\dfrac{u^2}{4g}$ .

Therefore, $\boxed {H_1=H_2}$

Projectile Motion Comparison - 2

1 solution

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