SOLUTION: Trajectory of a Projectile the height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2+29.4t+34.3, where (t) represents the end of the ramp. Deter

Question 1068200: Trajectory of a Projectile the height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2+29.4t+34.3, where (t) represents the end of the ramp.
Determine the time interval that the ball is above 50 m.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Trajectory of a Projectile the height in meters of a ball released from a ramp is given by the function h(t) = -4.9t^2+29.4t+34.3, where (t) represents the end of the ramp.
Determine the time interval that the ball is above 50 m.
:
h(t) = 50 meters
-4.9t^2 + 29.4t + 34.3 = 50
-4.9t^2 + 29.4t + 34.3 - 50 = 0
-4.9t^2 + 29.4t - 15.7 = 0
Use the quadratic formula to find t, I got
t = .59 seconds, 50 ft on the way up
and
t = 5.41 seconds, 50 ft on the way down
therefore
5.41 - .59 = 4.82 seconds it was at or above 50 ft
:
Graphically,the green line is 50 ft

You can see the time interval above 50 m is almost 5 seconds
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