Q.8.A ball is thrown with speed v0 at angle θ with respect to the horizontal. It is thrown from a point that is a distance l from the base of a cliff that has a height also equal to l. What should θ and v0 be so that the ball hits the corner of the cliff moving horizontally
Answers
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Explanation:
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Given:
The speed of ball = V₀
Angle = θ
Distance from the base = L
Height of the cliff = L
To find: the value of v₀ and θ
Solution:
From the equation of the projectile motion, Range = v₀²Sin2θ/g
L = v₀²Sin2θ/g
v₀² = Lg/ Sin2θ
v₀ = √(Lg/Sin2θ)
Now, as given in question, Range = Height
Maximum height = v₀²Sinθ/2g
v₀²Sin2θ/g = v₀²Sin²θ/2g
Sin2θ = Sin²θ/2
2SinθCosθ = Sin²θ/2
2Cosθ = Sinθ/2
4 = Sinθ/Cosθ
4 = Tanθ
θ = Tan⁻¹4
Therefore, the value of v₀ = √(Lg/Sin2θ)
, the value of θ = Tan⁻¹4