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# A ball is thrown vertically into the air with an initial velocity of 19m/s. What height will it travel upwards before it starts to reverse its direction?

Initial velocity u = 19m/s Final velocity v = 0m/s at the top Acceleration a = G = - 9.8m/s2 We need the distance s. Equation that contains these variables but not time is V2 = u2 + 2as U2 = - 2as U2 / 2a = - s Putting values in 19^2 / 2 * - 9.8 = - s 361/19.6 = s S=18.41m Ignore the syntax capital lower case etc. Hope this helps

H Ali
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686 students helped

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