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Question

A dynamic blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a height of after t sec. The velocity
of the rock when it is 256 m above the ground on the way up is

Moderate

A

98 m/s
B

96 m/s
C

104 m/s
D

48 m/s

Solution

$\begin{array}{l} v = \frac{d s}{d t} = 160 - 32 t . We now find values of t for which \\ s (t) = 256 . So \end{array}$
$\begin{array}{l} \begin{array}{cc} 160 t - 16 t^{2} = 256 \\ \Rightarrow & 16 (t^{2} - 10 t + 16) = 0 \\ \Rightarrow & (t - 2) (t - 8) = 0) \\ \Rightarrow & t = 2, t = 8 \\ So & v (2) = 160 - 32.2 = 96 \\ v (8) = 160 - 256 = - 96 \end{array} \\ \underline{\underline{So}} the velocity on the way up in 96 m / s . \end{array}$

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