Applications Of Derivatives Question 71
The maximum height is reached in 5 seconds by a stone thrown vertically upwards and moving under the equation s = ut - 4.9 $ t^{2} $ , where s is in metre and t is in second. The value of u is
Options:
A) $ 4.9,\text{m}/\sec$
B) $ 49,\text{m}/\sec$
C) $ 98,\text{m}/\sec$
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ \frac{dy}{dt}=-\frac{16}{6}=-\frac{8}{3}cm/\sec $ or $ s=ut-4.9t^{2} $
therefore $ \frac{ds}{dt}=u-9.8t=v $ When stone reached the maximum height, then $ v=0 $
therefore $ u-9.8t=0\Rightarrow u=9.8t $ But time $ t=5 $ sec So the value of $ u=9.8\times 5=49.0 $ m/sec
Hence initial velocity $ =49 $ m/sec.
BETA
