Question

monizontal range of projectile is fourtimrz of the maximum height allowed the angle of projectipnwith thr horizonatal will be​

Answer 1

$\maltese$Given:-

• Maximum Range (R) = 4 × Maximum Height (H)

$\maltese$To find:-

• Angle of projection $(\theta)$

$\maltese$Answer:-

As we know that,

$▪ \sf \: R = \dfrac{u {}^{2}sin(2 \theta) }{g}$

$▪ \sf \: H = \dfrac{u {}^{2}sin {}^{2}( \theta) }{2g}$

According to the question

$\sf \: R = 4H$

$\longrightarrow \sf \: \dfrac{ {u}^{2}sin(2 \theta) }{g} = 4 \times \dfrac{ {u}^{2} {sin}^{2} ( \theta) }{2g}$

$\longrightarrow \sf \: \dfrac{ \cancel{{u}^{2}}sin(2 \theta) }{ \cancel{g}} = 4 \times \dfrac{ \cancel{{u}^{2}} {sin}^{2} ( \theta) }{2 \cancel{g}} \:$

$\longrightarrow \sf sin(2 \theta) = 2 \times sin( \theta) \times sin( \theta)$

$\sf \: \longrightarrow 2 \times sin( \theta) \times cos( \theta) = 2 \times sin( \theta) \times sin( \theta)$

$\sf \: \longrightarrow \cancel{2} \times \cancel{sin( \theta)} \times cos( \theta) = \cancel{2} \times \cancel{sin( \theta)} \times sin( \theta)$

$\longrightarrow \sf \dfrac{sin( \theta)}{cos( \theta)} = 1$

$\longrightarrow \sf \tan( \theta) = 1$

From trigonometry ratio table, we know that,

$\sf \: tan( {45}^{o} ) = 1$

So,

$\longrightarrow \sf tan( \theta) = tan( {45}^{o} )$

$\longrightarrow \underline{ \underline{ \sf{ \green{ \theta = {45}^{o} }}}}$