How Do You Calculate the Time a Ball Stays Above a Certain Height?

AI Thread Summary
To calculate the time a ball stays above a certain height, the kinematic equation y = y0 + vy0t + 0.5at^2 is used, where y0 is the initial height, vy0 is the initial velocity, and a is the acceleration due to gravity. In this scenario, the ball is thrown upwards at 10 m/s from a height of 2 m, and the goal is to determine the time it remains above 3 m. The correct answer is 1.83 seconds, but the method described involved a lengthy calculation of total flight time and subtracting intervals below 3 m. A more efficient approach would involve solving a quadratic equation derived from the kinematic equation directly. This method simplifies the process and yields the same result more quickly.
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I have my physics exams on Friday so was just revising some of the stuff I covered earlier in the year. I got this question:

"A ball is thrown vertically upwards with a speed 10ms^-1 from a point 2m above horizontal ground. Calculate the length of time for which the ball is 3m or more above the ground."

I did end up getting the correct answer of 1.83s, but I took an extremely long-winded route. I calculated the total time the ball would be in flight, then subtracted the initial part under 3m and the final part under 3m, which I calculated using the simple equations of motion.

I seem to remember earlier on in the year using a much easier and quicker method to get the answer, but I have forgotten it, and can't seem to find it in my notes. How would you go about solving this question?
 
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y = y_{0} + v_{y0}t + \frac{1}{2}at^2

is the general kinematic equation you want to use,
where in your case:

solve the resulting quadratic equation for t
 
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