Question

7. A particle is projected at an angle 1p
45°. The relation between range
and maximum height attained by
the particle is *

Answer 1

Answer:

$\boxed{\mathfrak{R = 4H_{max}}}$

Explanation:

Angle of projection of projectile is given as:

$\rm \theta = 45\degree$

Maximum height of projectile:

$\boxed{ \bold{H_{max} = \dfrac{ {u}^{2} {sin}^{2} \theta }{2g} }}$

Horizontal range of projectile:

$\boxed{ \bold{R = \dfrac{ {u}^{2}sin2 \theta }{g} }}$

u → Initial velocity of projectile

g → Acceleration due to gravity

So, relation between range and maximum height attained by particle is:

$\rm \implies \dfrac{H_{max}}{R} = \dfrac{ \dfrac{ \cancel{ {u}^{2}} {sin}^{2} \theta }{2 \cancel{g}} }{ \dfrac{ \cancel{ {u}^{2}}sin2 \theta }{ \cancel{g}} } \\ \\ \rm \implies \dfrac{H_{max}}{R} = \dfrac{ \dfrac{ {sin}^{ \cancel{2}} \theta }{2} }{ 2 \cancel{sin \theta} cos \theta } \\ \\ \rm \implies \dfrac{H_{max}}{R} = \dfrac{ sin \theta}{2 \times 2cos \theta } \\ \\ \rm \implies \dfrac{H_{max}}{R} = \dfrac{ tan \theta}{4 } \\ \\ \rm \implies \dfrac{H_{max}}{R} = \dfrac{ tan 45 \degree}{4 } \\ \\ \rm \implies \dfrac{H_{max}}{R} = \dfrac{ 1}{4 } \\ \\ \rm \implies R = 4H_{max}$

Answer 2

Answer:

Maximum height, H = (u²sin²θ)/2g

Given the angle of projection is 45°,

∴ H= [u²(1/√2)²]/2g= u²/4g = 1/4 (u²/g)---(1)

Range , R = (u²sin2θ)/g

∴R = (u²sin90°)/g = u²/g

Hence, (1)⇒ H = R/4