JEE Main 12 Apr 2019 Paper 1 Solved Paper

Section: Physics

Question : 26 of 90

Marks: +1, -0

The trajectory of a projectile near the surface of the earth is given as y=2 x−9x2 . If it were launched at an angle θ0 with speed v0 then ( g=10ms−2).

[12 Apr 2019 Shift 1]

θ0=cos−1 (
1
√5
) and v0=
5
3
ms−1
θ0=cos−1 (
2
√5
) and v0=
5
3
ms−1
θ0=sin−1 (
2
√5
) and v0=
5
3
ms−1
θ0=sin−1(
1
√5
) and v0=
5
3
ms−1

Validate

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