Applications Of Derivatives Question 32
Question: A stone thrown vertically upward satisfies the equation $ s=64t-16t^{2} $ , where s is in meter and t is in second. What is the time required to reach the maximum height-
Options:
A) 1s
B) 2s
C) 3s
D) 4s
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given equation is $ s=64t-16t^{2} $
$ \therefore $ On differentiating w.r.t. t, we get $ \frac{ds}{dt}=64-32t $ Put $ \frac{ds}{dt}=0 $ for maximum height
$ \Rightarrow 64-32t=0 $
$ \Rightarrow t=2 $ Now, $ \frac{d^{2}s}{dt^{2}}=-32 $ At $ t=2,\frac{d^{2}s}{dt^{2}}=-32 $ Since, $ {{( \frac{d^{2}s}{dt^{2}} )}_{t=2}}<0 $
$ \therefore $ Required time = 2 second
BETA
