Applications Of Derivatives Question 374
Question: The equations of motion of two stones thrown vertically upwards simultaneously are $ s=19.6,t-4.9,t^{2} $ and $ s=9.8,t-4.9,t^{2} $ respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be
Options:
h/3
2h
h
0
Show Answer
Answer:
Correct Answer: D
Solution:
The time taken by first stone to secure maximum height = $ t=\frac{u}{g}=\frac{19.6}{9.8}=2sec $ .
The time taken by second stone to reach maximum height is, $ t=\frac{9.8}{9.8}= $ 1sec. Therefore, in 2 sec, second stone will return to the ground.
Hence height $ s=ut+\frac{1}{2}at^{2}+6 $ .
BETA
