A body is projected with a velocity v at an angle of projection - with the horizontal- The direction of velocity of the body makes angle 30- with the horizontal at t-2-s and then after further 1-s- it reaches the maximum height- Then-g-10-m-s-2-

Time of ascent =2+1=3 sec ⇒v sinθg=3 ⇒ v sinθ=30 Now, tanβ=v_yv_x ⇒tan 30^∘=v sinθ gtv cosθ ⇒1√(3)=30 20v cosθ⇒ v cosθ=10√(3) ∴ v=√(v_x^2+v_y^2)=√((10√(3))^2+(3 ...

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Byju's Answer

Standard XII

Physics

Applications of Horizontal and Vertical Components

A body is pro...

Question

A body is projected with a velocity $v$ at an angle of projection $θ$ with the horizontal. The direction of velocity of the body makes angle $30^{\circ}$ with the horizontal at $t = 2 s$ and then after further $1 s$ , it reaches the maximum height. Then
$(g = 10 {m/s}^{2})$

v = 20 \sqrt{3} m/s

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θ = 60^{\circ}

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θ = 30^{\circ}

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v = 10 \sqrt{3} m/s

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Solution

The correct options are
A $v = 20 \sqrt{3} m/s$
B $θ = 60^{\circ}$
Time of ascent $= 2 + 1 = 3 sec$
$\Rightarrow \frac{v sin θ}{g} = 3$
$\Rightarrow v sin θ = 30$
Now,
$tan β = \frac{v_{y}}{v_{x}}$
$\Rightarrow tan 30^{\circ} = \frac{v sin θ - g t}{v cos θ}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{30 - 20}{v cos θ} \Rightarrow v cos θ = 10 \sqrt{3}$
$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(10 \sqrt{3})^{2} + (30)^{2}} = 20 \sqrt{3} m/s$
and $tan θ = \frac{v_{y}}{v_{x}} = \sqrt{3} \Rightarrow θ = 60^{\circ}$

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A body is projected with a velocity v at an angle of projection θ with the horizontal. The direction of velocity of the body makes angle 30∘ with the horizontal at t=2 s and then after further 1 s, it reaches the maximum height. Then (g=10 m/s2)

The correct options are A v=20√3 m/s B θ=60∘Time of ascent =2+1=3 sec ⇒vsinθg=3 ⇒vsinθ=30 Now, tanβ=vyvx ⇒tan30∘=vsinθ−gtvcosθ ⇒1√3=30−20vcosθ⇒vcosθ=10√3 ∴v=√v2x+v2y=√(10√3)2+(30)2=20√3 m/s and tanθ=vyvx=√3⇒θ=60∘

A body is projected with a velocity $v$ at an angle of projection $θ$ with the horizontal. The direction of velocity of the body makes angle $30^{\circ}$ with the horizontal at $t = 2 s$ and then after further $1 s$ , it reaches the maximum height. Then
$(g = 10 {m/s}^{2})$