Question

a body projected with a velocity 30m/s at an angle of 30° with the vertical.find the maximum height attained by body​

Answer 1

Answer:

The maximum height attained by the body is 33.75 m

Explanation:

Given :

A body projected with a velocity 30 m/s at an angle of 30° with the vertical.

To find :

the maximum height attained by the body

Solution :

The maximum height of the projectile is given by,

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

where

u denotes the velocity of projection

θ denotes the angle made by the projected body with the horizontal

g denotes the acceleration due to gravity ( = 10 m/s² )

Here,

• 30° is the angle with the vertical.

Hence, angle made by the projected body with the horizontal,

θ = 90° - 30°

θ = 60°

• u = 30 m/s

Substituting the values,

$\sf H=\dfrac{30^2 \times sin^2 (60^{\circ})}{2(10)} \\\\ \sf H=\dfrac{900 \times \bigg( \dfrac{\sqrt{3}}{2}\bigg)^2}{20} \\\\ \sf H=\dfrac{900 \times \dfrac{3}{4}}{20} \\\\ \sf H=\dfrac{2700}{80} \\\\ \sf H=\dfrac{270}{8} \\\\ \sf H=33.75 \ m$

Therefore, the maximum height attained by the body is 33.75 m

Answer 2

$\\ \longrightarrow\small{ \sf H_{(Max)} = \dfrac{ {u}^{2} { \sin}^{2} \theta }{2g} }$

$\\ \longrightarrow \small{ \sf H_{(Max)} = \dfrac{ {(30)}^{2} { \sin}^{2} {30}^{0} }{2g} }$

$\\ \longrightarrow \small{ \sf H_{(Max)} = \dfrac{ {(30)}^{2} { \bigg(\dfrac{1}{2} \bigg) ^{2} } }{2(10)} }$

$\\ \longrightarrow \small{ \sf H_{(Max)} = \dfrac{30 \times 30 \times 20 }{4} }$

$\\ \longrightarrow \small{ \sf H_{(Max)} = 4500m }$