Question

a body is projected with an initial velocity of 20m-1 at an angle of30°with the horizontal caluculate
a)maximum height
b)time taken to reach the maximum height and
c)horizontal range​

Answer 1

Answer:

answer

Explanation:

Here u=30ms

−1 , Angle of projection, θ=90−30=60∘

Maximum height,

H= 2gu 2 sin 2 θ = 2×10302 sin260° = 20900 × 43 = 4135m Time of flight, T= g2usinθ = 102×30×sin60∘

=33sHorizontal range = R= gu 2 sin2θ = 1030×30×2sin60∘ cos60∘ =453m

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