A ball is thrown upwards from the top of a to | vedclass

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(C) Let the upward direction be positive. The initial position of the ball is $y_0 = 40\,m$ and the final position when it hits the ground is $y = 0\,

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$4$

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(C) Let the upward direction be positive. The initial position of the ball is $y_0 = 40\,m$ and the final position when it hits the ground is $y = 0\,m$. The displacement is $s = y - y_0 = 0 - 40 = -40\,m$.The initial velocity is $u = +10\,m/s$ and the acceleration is $a = -g = -10\,m/s^2$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$-40 = 10t + \frac{1}{2}(-10)t^2$$-40 = 10t - 5t^2$Dividing by $-5$:$8 = -2t + t^2$$t^2 - 2t - 8 = 0$Factoring the quadratic equation:$(t - 4)(t + 2) = 0$Since time cannot be negative,we take $t = 4\,s$.

$A$ ball is thrown upwards from the top of a tower $40\,m$ high with a velocity of $10\,m/s$. Find the time when it strikes the ground (in seconds). Take $g = 10\,m/s^2$.

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