SOLUTION: Grade 10- The height of a ball thrown upward after a given amount of time is h = - 4.9t ^ 2 + 29.4t + 1 where h represents the height of the ball in metres , and t represents the t

Question 1178898: Grade 10- The height of a ball thrown upward after a given amount of time is h = - 4.9t ^ 2 + 29.4t + 1 where h represents the height of the ball in metres , and t represents the time elapsed since the ball has been thrown . Determine the maximum height of the ball and at what time this occurs. What is the maximum height reached by the ball? How long is the ball above a height of 40 m ? When does the ball hit the ground?
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
the vertex t value is at -b/2a or -29.4/-9.8 or 3 seconds
the maximum height is at h(3)=-44.1+88.2+1=45.1 m
solve for h=40
so -4.9t^2+29.4t+1=40
and -4.9t^2+29.4t-39=0
or 4.9t^2-29.4t+39=0
t=(1/9.8)(29.4+/-sqrt (99.96); sqrt term=10.0
t=1.98 sec and t=4.02 sec
it is above 40 m for 2.04 sec
it hits 0 at -4.9t^2+29..4t-1=0=4.9t^2-29.4t+1
the positive root is t=(1/9.8)*(29.4+sqrt883.86); sqrt term=29.73
t=6.03 sec

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.

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