SOLUTION: a firework is launched from the top of a 216ft building with an initial upward velocity of 150 ft/sec
h(t)=-16t^2+vt+h
t=time, h is initial height, v is the initial velocity
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Question 1026866: a firework is launched from the top of a 216ft building with an initial upward velocity of 150 ft/sec
h(t)=-16t^2+vt+h
t=time, h is initial height, v is the initial velocity in feet per second
1.what is the equation for this situation
2.when will the firework land if it does not explode
3.show a table for this situation so that it shows the height from time t=0 until it hits the ground
4.calculate the axis of symmetry
5.calculate the coordinates of the vertex
6.explain why negative values of t and h(t) do not make sense for this problem
7. graph the situation
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
a firework is launched from the top of a 216ft building with an initial upward velocity of 150 ft/sec
h(t)=-16t^2+vt+h
t=time, h is initial height, v is the initial velocity in feet per second
1.what is the equation for this situation
h(t) = -16t^2 + 150t + 216
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2.when will the firework land if it does not explode
When h(t) = 0
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3.show a table for this situation so that it shows the height from time t=0 until it hits the ground
Pick values for t and find h(t)
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4.calculate the axis of symmetry
t = -b/2a
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5.calculate the coordinates of the vertex
--> h(t) at t = -b/2a
=====================
6.explain why negative values of t and h(t) do not make sense for this problem
The projectile's motion starts at t = 0
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7. graph the situation
Graph the parabola. I wouldn't call that a situation.
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