SOLUTION: A soccer ball is kicked into the air and its height in meters above ground, h, recorded after t seconds. The height of the ball is described by the function h(t)= -3t^2+24t+1.
a)
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Question 254720: A soccer ball is kicked into the air and its height in meters above ground, h, recorded after t seconds. The height of the ball is described by the function h(t)= -3t^2+24t+1.
a) Determine when the ball reached its maximum height. (I think 1 sec)
b) Determine the ball's maximum height.
c) Determine the instantanneous velocity of the projectile at t= 2.8 sec.
d) Determine the instantaneous velocity of the projectile at t= 4.5 sec.What does the negative sign mean? ( That is is falling back to the ground?)
e) Determine the instantaneous velocity of the projectile at t=4 sec.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A soccer ball is kicked into the air and its height in meters above ground, h, recorded after t seconds. The height of the ball is described by the function h(t)= -3t^2+24t+1.
a) Determine when the ball reached its maximum height. (I think 1 sec)
Max height is at the vertex of the parabola.
at t = -b/2a
t = -24/-6 = 4 seconds
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b) Determine the ball's maximum height.
h(4) = -3*16 + 96 + 1
h(4) = 49 meters
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c) Determine the instantaneous velocity of the projectile at t= 2.8 sec.
h'(t) = -6t + 24
h'(2.8) = -16.8 + 24
h'(2.8) = 7.2 meters/sec (positive, going up)
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d) Determine the instantaneous velocity of the projectile at t= 4.5 sec.What does the negative sign mean? ( That is is falling back to the ground?)
h'(4.5) = -6*4.5 + 24
h'(4.5) = -3 meters/second (negative --> going back down)
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e) Determine the instantaneous velocity of the projectile at t=4 sec.
h'(4) = -6*4 + 24 = 0 (it stops going up, and starts back down)
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h' is the 1st derivative of height with respect to time. I don't know how to do this without derivatives.
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