Question

write the equation for the horizontal range covered by a projectile and specify when it will be maximum​

Answer 1

Answer:

Given:

A projectile has been thrown at an angle of θ with the horizontal.

To find:

Equation of horizontal range and stating the condition in which the range will be maximum.

Concept:

The Equation for horizontal range of the projectile is given as :

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

where , u => initial Velocity , and

g => acceleration due to gravity.

Condition when range is maximum:

For range to be maximum, the value of sin (2θ) has to 1 .

Hence :

∴ sin(2θ) = 1

=> 2θ = 90°

=> θ = 45°

So the angle of Projection has to be 45° for max range.

Answer 2

Range of Projectile

The maximum horizontal distance covered by the projectile is known as Range

Since,

$\boxed{ \boxed{ \tt{speed = \frac{d}{t} }}}$

• d is the distance covered

• t is the time taken

Implies,

$\tt{d = speed \times t}$

Velocity of a projectile along x - axis (horizontally) is

$\tt{speed = u \cos( \theta) - - - - - (1) }$

Time taken by a projectile is given by,

$\tt{t = \dfrac{2u \sin( \theta) }{g} - - - - (2)}$

From equations (1) and (2),we get :

$\tt{r = u \cos( \theta) \times \dfrac{2u \sin( \theta) }{g} } \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt{ r = \frac{ {u}^{2} \sin(2 \theta) }{g} }}}$

For maximum range,the maxima of sine of the angle would be 1

$\implies \: \tt{sin \theta = 1} \\ \\ \implies \: \underline{ \boxed{ \tt{ \theta = {90}^{ \circ} }}}$