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Question: A body thrown vertically upwards with an initial velocity \(u\) reaches maximum height in 6 seconds....

A body thrown vertically upwards with an initial velocity uu 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

Answer

11 : 1

Explanation

Solution

Time of ascent

=ug=6 6musec⁡⇒u=60 6mum/s= \frac{u}{g} = 6\mspace{6mu}\sec \Rightarrow u = 60\mspace{6mu} m/s

Distance in first second

hfirst=60−g2(2×1−1)=55 6mumh_{\text{first}} = 60 - \frac{g}{2}(2 \times 1 - 1) = 55\mspace{6mu} m

Distance in seventh second will be equal to the distance in first second of vertical downward motion

hseventh=g2(2×1−1)=5 6mumh_{\text{seventh}} = \frac{g}{2}(2 \times 1 - 1) = 5\mspace{6mu} m ⇒ hfirst/hseventh=11:1h_{\text{first}}/h_{\text{seventh}} = 11:1