A ball is thrown vertically upward with an initial velocity of 96 feet per secon

Question

A ball is thrown vertically upward with an initial velocity of 96 feet per second. the distance (s) in feet of the ball from the ground after (t) seconds is s (t) = 96t - 16t^2 a ) AT WHAT TIME WILL THE BALL HIT THE GROUND ? b ) FOR WHAT TIME ( t ) IS THE BALL MORE THAN 128 FEET ABOVE THE GROUND ?? c) WHEN WILL THE BALL REACH IT HIGHEST PEAK?? HOW HIGH IS IT ABOVE THE GROUND ??

Explanation / Answer

a When s(t) = 0 96t - 16t^2 = 0 16t(6-t) = 0 6 - t = 0 t = 6 (We are not considering t = 0, as this is when the ball is thrown. b) 96t - 16t^2 > 128 16t^2 - 96t + 128 < 0 Dividing by 16, t^2 - 6t + 8 < 0 (t - 2)(t-4) < 0 This means t-4 < 0 and t-2 > 0 (we never have the opposite condition, and 1 term must be positive and the other negative to have a negative number). t-4 < 0 and t-2 > 0, so t < 4 and t > 2 Thus, 2

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