Kinematics Question 415
Question: From a pole of height 10 m, a stone is thrown vertically upwards with a speed 5 m/s. The time taken by the stone, to hit the ground, it’s n times that taken by it to reach the highest point of it’s path. The value of n is $ [takeg=10m/s^{2}] $
Options:
A) 2
B) 3
C) 4
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
Given, $ H=10m $ , $ u=5m/s,g=10m/s^{2} $
Speed on reaching ground $ v=\sqrt{u^{2}+2gh} $ Now, v = u + at Now, $ v=\sqrt{u^{2}+2gh}=-u+gt $
Time taken to reach highest point is $ t=\frac{u}{g} $ ,
$ \Rightarrow t=\frac{u+\sqrt{u^{2}+2gH}}{g}=\frac{nu}{g}\text{(from question)} $
$ \Rightarrow 2gH=n( n-2 )u^{2} $
$ \Rightarrow m( n-1 )=\frac{2gH}{u^{2}}=\frac{2\times 10+10}{5\times 5}=8\Rightarrow n=4 $
