A body projected from the ground reaches a point X in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point X from the ground is (acceleration due to gravity =10 ms -2 )

Question

A body projected from the ground reaches a point X in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point X from the ground is (acceleration due to gravity =10 ms -2 ) (A) 30 m (B) 60 m (C) 80 m (D) 90 m

Tardigrade Admin · Accepted Answer

Given, acceleration due to gravity (g)=10 ms -2 According to the question, Total time of flight (t)=(2 u/g) where, u is initial vertical velocity. or, 9=(2 u/g) or,u=(9 g/2) or, u=(9 × 10/2) or, u=45 m / s Since, in covering the vertical distance, g becomes negative, ⇒ h=u t-(1/2) g t2 Or, h=45 ×(3)-(1/2) × 10 ×(3)2 Or, h=135-45 Or, h=90 m

Q. A body projected from the ground reaches a point X in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point X from the ground is (acceleration due to gravity =10ms−2 )

30 m

60 m

80 m

90 m

Q. A body projected from the ground reaches a point $X$ in its path after $3$ seconds and from there it reaches the ground after further $6$ seconds. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 m s^{- 2}$ )