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Kinematics in one dimension question

Hello everyone. I need some help:

"A ball is thrown vertically upwards from a cliff which is 50m high. The initial velocity of the ball is 25 m/s. Calculate the time taken to reach the bottom"

I re-arranged the equation s=ut+0.5at^2 to calculate time, and got a value around 9 seconds... but the book says that the answer is 6.63...

Can anyone help?
Original post by Electrogeek
Hello everyone. I need some help:

"A ball is thrown vertically upwards from a cliff which is 50m high. The initial velocity of the ball is 25 m/s. Calculate the time taken to reach the bottom"

I re-arranged the equation s=ut+0.5at^2 to calculate time, and got a value around 9 seconds... but the book says that the answer is 6.63...

Can anyone help?


Take upwards to be the positive direction, and the top of the cliff to be 0 displacement. so u=25 a=-9.81 and s=-50. when you put this into the s=ut + 0.5at^2 equation you obtain a quadratic that can be solved using the quadratic formula. You get -4.9t^2 +25t +50 = 0 which solves for t=6.63 and t=-1.54 . then take the positive value for t
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Electrogeek
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It turns out that I didn't take s as -50... Thanks for the help. :smile:

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