Question

Relation between time of flight and range

Answer 1

Answer:

$\large\bold\red{ R = T(u \cos \theta )}$

Explanation:

In a projectile motion,

• When a body of some mass is thrown at an angle $\bold{\theta}$ with some initial velocity u,
• It follows a path and fall at some distance in certain time interval.
• That time taken to complete it's journey is called Time of Flight (T).

And,

The time of flight is given as,

$T = \frac{2u \sin( \theta ) }{g} \: \: \: \: \: .........(i)$

Also,

• The maximum horizontal distance travelled is called the Range.

And,

The Range is given as,

$R = \frac{ {u}^{2} \sin(2 \theta ) }{g}$

But,

We know that,

• $\sin(2 \theta ) = 2 \sin( \theta ) \cos( \theta )$

Therefore,

Putting the values,

We get,

$= > R = \frac{ {u}^{2}2 \sin( \theta ) \cos( \theta ) }{g} \\ \\ = > R = ( \frac{2u \sin\theta }{g}) (u \cos \theta )$

But ,

from (i),

Substituting the values,

We get,

$= > \large\bold{R = T(u \cos \theta )}$