Question

49
1. A stone is thrown up from the earth's surface with an
initial velocity of 28 m.s-1, making an angle of 30°
with the horizontal. Find the maximum height attained
and the range of the projectile motion of the stone.​

Answer 1

Answer

Given

• Intial velocity (u) = 28 m/s
• Angle of projection (θ) = 30°

To find

• Maximum height (H)
• Horizonatal Range (R)

Solution

$\bf\red{Calculating \:Maximum \:height (H)}$

• u = 28 m/s
• θ = 30°
• g = 9.8 m/s²

Substituting the value in formula :-

$\boxed{\sf H = \frac{u^2sin^2\theta}{2g}}$

$\sf H = \frac{(28)^2sin^2 30}{2\times 9.8}$

$\sf H = \frac{(28)^2 \times sin^2 30}{2\times 9.8}$

$\sf H = \frac{784 \times \frac{1}{4} }{19.6}$

$\sf H = \frac{ \cancel{ 196}}{\cancel{ 19.6} }$

$\sf H = 10m$

$\boxed{\underline{\rm\purple{Maximum \:height = 10 m }}}$

$\bf\red{Calculating \:Horizontal \:Range (R)}$

• u = 28 m/s
• θ = 30°
• g = 9.8 m/s²

Substituting the value in formula :-

$\boxed{\sf R = \frac{u^2sin2\theta}{g}}$

$\sf R = \frac{(28)^2 \times sin(2 \times 30)}{9.8}$

$\sf R = \frac{784 \times \frac{ \sqrt{3} }{2} }{9.8}$

$\sf R = \frac{ \cancel{ 392} \times \sqrt{3} }{\cancel{ 9.8}}$

$\sf R = 40 \times \sqrt{3}$

$\sf R = 69.28$

$\boxed{\underline{\rm\purple{Horizontal \:range= 69.28 m }}}$