Question

a ball is thrown upwards with an initial velocity of 15.5ms-1. for how long is the ball higher than 10.5 meters

Answer 1

from formula s= ut-1/2 gt^2
put s and u you will get two values of t ,difference between t is your answer.

Answer 2

why the difference in t is the answer ?

Answer 3

because ball would be two times as position 10.5 m once when it would go up and second when it come down
s= ut-1/2 gt^2 tells us these two values of time
so in between this time interval ball would be higher than 10.5 meters.

Answer 4

Using the formula :
y = Vo *t - ( 0.5 * 9.81 m/s^2 * t^2)

y = distance in y-axis to where time begins
Vo = initial velocity
t = time. There is the time from where ball goes above 10.5m and peaks, and then the time for it to come back to 10.5m before going to ground

10.5m = (15.5m/s) * t - ( 0.5 * 9.81m/s^2 * t^2)
solving for t , there is two times , t = 0.9835 seconds
second time before goes below 10.5m , t = 2.1765 seconds

SO the time above 10.5m is t2 - t1 :
2.1765s - 0.9835s = 1.193 seconds for flight time above 10.5 meter
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See this link for calculation and a graph of the flight and the two points above 10.5m. I substituted in an x fr the time t, but same results:

http://www.wolframalpha.com/input/?i=10.5+%3D+15.5x-%280.5*9.81*x^2%29