A ball is thrown vertically upward from the top of a tower with speed 19.6 m s⁻¹ and strikes the ground after 6 s. The maximum height from ground reached by the ball is (k/5) m. Take g=9.8 m s⁻². Find k.
Physics · Kinematics Part-1 (one dimension) · Motion under gravity
JEE Mains

Question

A ball is thrown vertically upward from the top of a tower with speed 19.6 m s⁻¹ and strikes the ground after 6 s. The maximum height from ground reached by the ball is (k/5) m. Take g=9.8 m s⁻². Find k.

Physics · Kinematics Part-1 (one dimension) · Motion under gravity

G1 Physics · Accepted Answer

Take launch point as y=0. Ground is reached in 6 s: y=ut-gt²/2=19.6(6)-4.9(36)=-58.8 m, so tower height is 58.8 m. The ball rises an extra u²/(2g)=19.6 m, so maximum height above ground=78.4 m=k/5. Hence k=392.