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(C) Given the equation of motion: $s = 490t - 4.9t^2$.To find the maximum height,we differentiate $s$ with respect to $t$ and set the derivative to ze

Vedclass · Accepted Answer

$12250$

Vedclass · Answer

(C) Given the equation of motion: $s = 490t - 4.9t^2$.To find the maximum height,we differentiate $s$ with respect to $t$ and set the derivative to zero:$\frac{ds}{dt} = 490 - 9.8t$.Setting $\frac{ds}{dt} = 0$ for maximum height:$490 - 9.8t = 0$$9.8t = 490$$t = \frac{490}{9.8} = 50 	ext{ seconds}$.Now,substitute $t = 50$ into the original equation to find the maximum height $s$:$s = 490(50) - 4.9(50)^2$$s = 24500 - 4.9(2500)$$s = 24500 - 12250$$s = 12250$.

$A$ stone thrown upwards has its equation of motion $s = 490t - 4.9t^2$. Then the maximum height reached by it is:

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