A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms-1. After 10 s, it inclination with horizontal is β. The value of tanβ will be: (g = 10ms-2).
A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms-1. After 10 s, it inclination with horizontal is β. The value of tanβ will be: (g = 10ms-2).
We know that horizontal speed will remain constant
20 cos = v cos ……… (i)
Along y-axis
……. (ii)
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Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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Physics NCERT Exemplar Solutions Class 11th Chapter Eight 2025
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