A ball is tossed into the air at 40 feet per second from ahight of 6 feet. How l

Question

A ball is tossed into the air at 40 feet per second from ahight of 6 feet. How long will it take the ball to reach the ground? Use the formula s= -16t2 + vot + so , where vo = initialvelocity, so = initial altitude, and s = altitude in feet after tseconds. The approximate time it took was (   )seconds. (round to the nearest tenth.) A ball is tossed into the air at 40 feet per second from ahight of 6 feet. How long will it take the ball to reach the ground? Use the formula s= -16t2 + vot + so , where vo = initialvelocity, so = initial altitude, and s = altitude in feet after tseconds. The approximate time it took was (   )seconds. (round to the nearest tenth.)

Explanation / Answer

vo   40 since that is the initial velocity; vo is 6since the ball is thrown from a height of 6 feet. so our equation is s = -16t^2 + 40t + 6      we set thisequal to 0 and solve since when s = 0 the ballwill                                    hit the ground. 16t^2 - 40t - 6 = 0   we first divide through by2 8t^2 - 20t - 3 = 0    now we solve by quadraticformula t = [ 20 ± (400 + 96)] / 16    = [ 20 ± 22.27 ] / 16 since time cannot be begative, we take the positive root so t = [20 + 22.27] / 16        = 44.27 /16         =2.78      which means that the ball willhit the ground at 2.8 seconds hope this helps

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