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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 $t$ to find the velocity $v = \

Vedclass · Accepted Answer

$12250$

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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 $t$ to find the velocity $v = \frac{ds}{dt}$.$\frac{ds}{dt} = 490 - 9.8t$.At maximum height, the velocity is zero, so set $\frac{ds}{dt} = 0$.$490 - 9.8t = 0$.$t = \frac{490}{9.8} = 50 	ext{ seconds}$.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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