HELP

[heather2005]

1. A certain baseball hit straight up in the air, is at height represented by:
h(t) = 4 + 50t – 16t2 , where t is time in seconds after being hit and h(t) is in feet. The function is valid for t > 0, until the ball hits the ground.

a) Sketch the graph of this function.

b) How high is it at t = 0?

c) When does it hit the ground?

d) When, if ever, is it 30 feet high?

e) When, if ever, is it 90 feet high?

f) When does it reach the maximum height?

g) What is the maximum height that it reaches?

 

[heather2005]

B. (0,4)

C. H=0 -16t^2+50t+4
8t^2-25t-4=0

d. 30=-16t^2+50t+4
16t^2-50t+26
t=1/16(25+209^1/2

 

[masters]

1. A certain baseball hit straight up in the air, is at height represented by:
h(t) = 4 + 50t – 16t2 , where t is time in seconds after being hit and h(t) is in feet. The function is valid for t > 0, until the ball hits the ground.

a) Sketch the graph of this function.

\(\displaystyle h(t)=-16t^2+50t+4\)

The graph is that of a parabola which opens downward.

The vertex is at \(\displaystyle \left(-\frac{b}{2a}, h\left(-\frac{b}{2a}\right)\right)=(1.5625, 43.0625)\)

The y-intercept is 4

b) How high is it at t = 0?

At t(0) the height is 4 feet.

c) When does it hit the ground?

Solve the function for t using the quadratic formula. Use the positive value of t.

d) When, if ever, is it 30 feet high?

Solve \(\displaystyle -16t^2+50t+4=30\) and check for positive values of t.

e) When, if ever, is it 90 feet high?

Do the same as question (d). Solve \(\displaystyle -16t^2+50t+4=90\) and check for positive values of t

f) When does it reach the maximum height?

That would be the x-coordinate of the vertex found in question (a). About 1.6 seconds.

g) What is the maximum height that it reaches?

That would be the y-coordinate of the vertex found in question (a). About 43.1 feet.

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