A stone thrown upwards, has its equation of motion $s=490 t-4.9 t^2$. Then the maximum height reached by it,…
A stone thrown upwards, has its equation of motion $s=490 t-4.9 t^2$. Then the maximum height reached by it, is
24500
12500
12250
25400
Solution
Given that
$s=490 t-4.9 t^2$
On differentiating w.r.t. $t$, we get
$\frac{d s}{d t}=490-9.8 t$
A stone is reached the maximum height, when $\frac{d s}{d t}=0$
$\begin{array}{cc}
\Rightarrow & 490-9.8 t=0 \\
\Rightarrow & t=\frac{490}{9.8}=\frac{100}{2}=50
\end{array}$
$\therefore$ Maximum height at $t=50$
$\begin{aligned}
s & =490(50)-4.9(50)^2 \\
& =24500-12250 \\
& =12250
\end{aligned}$