Maximum height of a projectile

Maximum height of projectile thrown from ground is given by $\cfrac{u^2 \sin^2 \theta}{2g}$ and if the projectile is projected from a height $H$, then the maximum height attained by projectile during it’s flight is $H + \cfrac{u^2 \sin^2 \theta}{2g}$ as measured from the ground

So let’s see how we can quickly derive the maximum height from the equations of motion of a projectile

Maximum Height in Projectile Motion

Maximum height of a projectile, i.e., the maximum displacement in the y -direction is governed only by the kinematic equations in y – direction i.e. $v_y =$ $u_y -$ $gt$ & $y =$ $ut -$ $\cfrac{1}{2} gt^2$ or $v_y^2 = u_y^2 – 2gy$

Now, what is the vertical component of velocity or speed in vertical direction at maximum height? Well, it is zero at that point i.e. $v_y = 0$ at maximum height

So $0 = u_y^2 – 2gH$ or $H = \cfrac{u_y^2}{2g}$ $=\cfrac{u^2 \sin^2 \theta}{2g}$

The horizontal range and maximum height of a projectile are equal - When?

Maximum height of a projectile, $H$ $=\cfrac{u^2 \sin^2 \theta}{2g}$, where once again $u$ is the initial speed, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity

Horizontal Range of a projectile, $R$ $=\cfrac{2u^2 \ sin \theta \cos \theta}{g}$

$H=R$ when $\cfrac{u^2 \sin^2 \theta}{2g}$ $=\cfrac{2u^2 \ sin \theta \cos \theta}{g}$

or when, $\tan \theta = 4$ or when the angle of projection, $\theta =$ $\tan^{-1} 4$

With that, now let’s explore the trajectory of a projectile, i.e. establish expression $y = f(x)$ for the path traced by the projectile

Projectile Motion | Important Questions | JEE PYQs

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