Question 80218
h(t) = rt - 4.9t^2

In how many seconds after it is thrown upward from the ground with an initial
speed of 34.3m/s will the object be 49m above the ground? 
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h(t) is the height in meters of the object at time "t" seconds.
r is the initial velocity of the object(r=34.3 meters/sec).
-4.9 is (1/2)g where g is the acceleration of the object due to gravity.
Your formula assumes the object started up from ground level.
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Your problem: Substitute to get:
49 = 34.3t-4.9t^2
Rearrange to get:
4.9t^2-34.3t+49=0
Solve for t:
t=[34.3+-sqrt(34.3^2-4*4.9*49]/{2*4.9)
t=35.8 seconds or t=32.8 seconds
The object will be at 49 meters first at t=32.8 seconds then at 35.8 seconds.
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Cheers,
Stan H.