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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.

Toolbox:
• 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.
edited Apr 28, 2014 by meena.p