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A missile fired from ground level rises $x$ metres vertically upwards in $t$ seconds and $x=100t-\large\frac{25}{2}t^{2}$ find the maximum hight reached .

Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

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  • If $s=f(t)$ is the distance function, representing the distance 's' travelled by a particle in time t, then the velocity and acceleration functions are $v=\large\frac{ds}{dt}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}=f''(t)$
  • When a particle starts from rest,velocity v and time t are 0. When a particle is thrown up, it reaches maximum height at which $v=0$ and then falls back to earth. When a moving particle comes rest, $v=0$
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$x=100t-\large\frac{25}{2}$$t^2$
The maximum height is obtained on substituting $t=4$ in (i)
$x=100(4)-\large\frac{25}{2} $$\times 16$
$\quad=400-200$
$\quad=200\;m$
Hence A is the correct answer.
answered Jul 22, 2013 by meena.p
edited Apr 28, 2014 by meena.p
 

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