The maximum height is given by, H=2gu2sin2θ.
The given data is
θ1=30∘θ2=60∘u1=u2
Therefore,
(H2)max(H1)max=u22sin2θ2u12sin2θ1=(sin60∘sin30∘)2=(2321)2=31
Two projectiles are projected at 30∘ and 60∘with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:
Held on 10 Apr 2023 · Verified 6 Jul 2026.
3:1
1:3
2:3
1:3
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