Question

02. A ball more
moving is thrown at an angle of 60° with
the horizontal.with initial speed of 30m/s . Calculate
Time of flight of the ball
(Ans: 5.3 sec)
​

Answer 1

$\underline{\rm\red{Question :}}$

A ball is thrown at an angle of 60° with horizontal with initial speed of 30 m/s. Calculate time of flight of ball.

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$\underline{\rm\red{Answer :-}}$

$\underline{\rm\pink{Given :}}$

$\rm u = 30 m/s$

$\rm \theta = 60\degree$

$\rm g = 9.8 m/s^2$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is angle of projection.

$\longrightarrow$g is acceleration due to gravity.

$\underline{\rm\pink{To\: find : }}$

Time of filght $\longrightarrow$ T

$\underline{\rm\pink{Formula \:used : }}$

$\boxed{\underline{\bf\purple{T = \dfrac{2usin\theta}{g}}}}$

$\underline{\rm\red{Solution :}}$

$\implies$$\rm u = 30 m/s$

$\implies$$\rm \theta = 60\degree$

$\implies$$\rm g = 9.8 m/s^2$

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Substituting the value in formula -

$\rm T = \dfrac{2usin\theta}{g}$

$\implies$$\rm T = \dfrac{2 \times 30 \times sin60}{g}$

$\implies$$\rm T = \dfrac{\cancel 2 \times 30 \times \dfrac{\sqrt{3}}{\cancel 2}}{g}$

$\implies$$\rm T = \dfrac{\sqrt{3} \times 30}{9.8}$

$\implies$$\rm T = 5.3 s$

$\boxed{\underline{\rm\purple{Time \: of \: Flight = 5.3 sec}}}$

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$\underline{\rm\red{Additional \:information - }}$

Horizontal range -

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

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Maximum Horizontal Range -

$\implies$$\boxed{\rm R_{max} = \dfrac{u^2}{g}}$

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Maximum height -

$\implies$$\boxed{\rm H = \dfrac{u^2sin^2\theta}{2g}}$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is angle of projection.

$\longrightarrow$g is acceleration due to gravity.

$\longrightarrow$R is range.

$\longrightarrow$H is maximum height.

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