Name: Parameters of Motion and Trajectory of Projectile
Uploaded: 2019-07-19
Duration: 10 min 13 s
Description: Mathematics is a great tool to analyse any motion. The expressions for the height, range and time of flight of projectile motion can be derived using math. L...

Question

Establish the expression for the range and maximum height attained by a projectile. Prove that speed remaining constant the range is same for two angles of projection.

Accepted Answer

Maximum height

The maximum vertical distance travelled by the projectile during the journey is called maximum height.

For the vertical part of the motion

${v_{y}}^{2} = {u_{y}}^{2} + 2 a_{y} s$

Here

$u_{y} = u \sin θ a = - g s = h_{m a x}$

And at maximum height, $v = 0$

$\Rightarrow 0 = {(u \sin θ)}^{2} + 2 (- g) (h_{m a x})$

$\Rightarrow h_{m a x} = \frac{u^{2} \sin^{2} θ}{2 g}$

Horizontal range

The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R).

This is found easily since the horizontal component of initial velocity remains the same.

Range = Horizontal component of velocity $\times$ Time of flight

$R = u \cos θ \times T$

As we know, Time of flight is given by

$T = \frac{2 u \sin θ}{g}$

So, Range becomes

$R = u \cos θ \times \frac{2 u \sin θ}{g}$

$\Rightarrow R = \frac{2 u^{2} \sin θ \cos θ}{g} \Rightarrow R = \frac{u^{2} \sin 2 θ}{g} [∵ \sin 2 θ = 2 \sin θ \cos θ]$

Now, using the expression for horizontal range which is given by

$R = \frac{u^{2} \sin 2 θ}{g}$

where $θ$ is the angle of projection and is less than $90^{\circ}$

We know,

$\sin (180^{\circ} - 2 θ) = \sin 2 θ \Rightarrow \sin 2 (90^{\circ} - θ) = \sin 2 θ \Rightarrow \sin 2 α = \sin 2 θ [w h e r e α = 90^{\circ} - θ]$

Now,

$R_{\circ} = \frac{u^{2} \sin 2 θ}{g}$

and $R_{α} = \frac{u^{2} \sin 2 α}{g}$

Since $\sin 2 α = \sin 2 θ ∴ R_{\circ} = R_{α}$

Thus, the horizontal range is same for two same angles, and it is to be noticed that speed of projection is also taken same.

Hence, proved.

Establish the expression for the range and maximum height attained by a projectile. Prove that speed remaining constant the range is same for two angles of projection.

Important Questions on Projectile Motion

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