A body is projected with a velocity v→=(3i^+4j^)ms-1. The maximum height attained by the body is  take g=10ms−2

v=(3i^+4j^)ms−1. The magnitude of velocity is v=(3)2+(4)2=5ms−1. The angle subtended by the velocity vector with the horizontal (x-axis) is given by tan⁡θ=VyVx=43which gives sin⁡θ=45.Now, hmax=v2sin2⁡θ2g=(5)2×4522×10=0.8mHence the correct choice is (a).