Q. A body thrown vertically upwards with an initial velocity $u$ reaches maximum height in 6 seconds. The ratio of the distances travelled by the body in the first second and the seventh second is
A) $1\ :1$
B) $11\ :1$
C) $1\ :2$
D) $1\ :11$
Explanation
(b) Time of ascent $= \frac{u}{g} = 6\ \sec \Rightarrow u = 60m/s$
Distance in first second $h_{first} = 60 - \frac{g}{2}(2 \times 1 - 1) = 55m$
Distance in seventh second will be equal to the distance in first second of vertical downward motion
$h_{seventh} = \frac{g}{2}(2 \times 1 - 1) = 5\ m\ \Rightarrow h_{first}/h_{seventh} = 11:1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.