A body is projected with a velocity u at an angle of projection θ with the horizontal. The direction of velocity of body makes angle 300 with the horizontal at t = 2 s and then after 1 s it reaches the maximum height then numerical addition u+θ=_______ ( u in ms−1 and θ in degree, g=10ms−2 and 3=1.73)

Time of ascent=2H=3 secusinθg=3⇒usinθ=30 And Tanβ=usinθ−gtgcosθtan30030−10×2ucosθucosθ=103, u=(103)2+(10)2=203 m/sAnd tanθ=usinθucosθ=30103=3⇒θ=600

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A body is projected with a velocity $u$ at an angle of projection $θ$ with the horizontal. The direction of velocity of body makes angle 30⁰ with the horizontal at t = 2 s and then after 1 s it reaches the maximum height then numerical addition $u + θ =$ _______ ( u in $m s^{- 1}$ and $θ$ in degree, $g = 10 m s^{- 2} a n d \sqrt{3} = 1.73$ )

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Detailed Solution

Time of ascent=2H=3 sec

$\frac{u \sin θ}{g} = 3 \Rightarrow u \sin θ = 30 A n d T a n β = \frac{u \sin θ - g t}{g \cos θ}$

$\tan 30^{0} \frac{30 - 10 \times 2}{u \cos θ}$

$u \cos θ = 10 \sqrt{3}, u = \sqrt{{(10 \sqrt{3})}^{2} + {(10)}^{2}} = 20 \sqrt{3} m / s$

And $\tan θ = \frac{u \sin θ}{u \cos θ} = \frac{30}{10 \sqrt{3}} = \sqrt{3} \Rightarrow θ = 60^{0}$

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