Question

an arrow is shot in air its time of flight is 5 sec and horizontal range is 200m inclination of an arrow with horizontal

Answer 1

Answer:

$\boxed{\mathfrak{Angle \ of \ inclination \ (\theta) = tan^{-1}(\frac{5}{8})}}$

Given:

Time of flight ($\sf T_f$) = 5 sec

Horizontal range (R) = 200 m

To Find:

Angle of inclination ($\sf \theta$) with horizontal

Explanation:

The total time for which projectile remains in flight is called it's time of flight.

$\boxed{ \bold{T_f = \frac{2u \: sin \theta}{g} }}$

The maximum horizontal distance travelled by the projectile during its flight is called the horizontal range of the projectile.

$\boxed{ \bold{R = \frac{2 {u}^{2} sin \theta.cos \theta}{g} }}$

In both the equation;

u $\rightarrow$ Initial velocity from ground

$\sf \theta$ $\rightarrow$ Angle of projectile with the horizontal

Dividing equation of time of flight by squaring it with horizontal range we get:

$\sf \implies \frac{T_f ^{2} }{R} = \frac{ (\frac{2u \: sin \theta}{ g} )^{2} }{ \frac{ 2 u ^{2} sin \theta.cos \theta}{g}}\\ \\ \sf \implies \frac{T_f ^{2} }{R} = \frac{ \frac{ 4 \cancel{u ^{2} } \: sin \theta^{ \cancel{2}} }{ g ^{ \cancel{2}} } }{ \frac{ 2 \cancel{u ^{2} } \cancel{sin \theta}.cos \theta}{ \cancel{g}} } \\ \\ \sf \implies \frac{T_f ^{2} }{R} = \frac{2 \: sin \theta}{g cos \theta} \\ \\ \sf \implies \frac{5 ^{2} }{200} = \frac{2tan \theta}{g} \\ \\ \sf \implies tan \theta = \frac{5 ^{2} \times g }{200 \times 2} \\ \\ \sf \implies tan \theta = \frac{25 \times 10}{400} \\ \\ \sf \implies tan \theta = \frac{250}{400} \\ \\ \sf \implies tan \theta = \frac{5}{8} \\ \\ \sf \implies \theta = {tan}^{ - 1} ( \frac{5}{8} )$

$\therefore$

Angle of inclination ($\sf \theta$) with horizontal = $\sf tan^{-1}(\frac{5}{8})$