Search any question & find its solution
Question:
Answered & Verified by Expert
A missile is fired from the ground level rises \(x\) meters vertically upwards in \(t \sec\), where \(x=100 t-\frac{25}{2} t^2\). The maximum height reached is
Options:
Solution:
1869 Upvotes
Verified Answer
The correct answer is:
\(200 \mathrm{~m}\)
Hint : \(x=100 t-\frac{25}{2} t^2\)
\(\frac{d x}{d t}=100-25 \mathrm{t}\) for maximum and minimum \(\frac{d x}{d t}=0 \Rightarrow t=4 \mathrm{sec}\)
\(\frac{d^2 x}{d t^2}=-25 < 0\)
\(\therefore\) at \(\mathrm{x}=4\) it will have maixmum height
\(x_{\max }=100 \times 4-\frac{25}{2} \times 16=400-200=200 \mathrm{~m}\)
\(\frac{d x}{d t}=100-25 \mathrm{t}\) for maximum and minimum \(\frac{d x}{d t}=0 \Rightarrow t=4 \mathrm{sec}\)
\(\frac{d^2 x}{d t^2}=-25 < 0\)
\(\therefore\) at \(\mathrm{x}=4\) it will have maixmum height
\(x_{\max }=100 \times 4-\frac{25}{2} \times 16=400-200=200 \mathrm{~m}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.