SOLUTION: a firework is launched from the top of a 50ft 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
Question 1009578: a firework is launched from the top of a 50ft 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.a table for this situation so that it shows the height from time t=0 unit 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 Boreal(15235) (Show Source):
You can put this solution on YOUR website! the equation is -16t^2+150t+50, the middle term being the upward velocity and the last term being the height it started at.
The vertex is at -150/-32=4.6875, and the height is 401.56 feet
Setting the equation equal to zero,
-16t^2+150t+50=0 or 16t^2-150t-50=0
t=(1/32) { 150 +/- sqrt (22500+3200)}, where sqrt (25700)=160.31
t=(1/32)310.31=9.697 sec or 9.7 seconds TIME TO LANDING
===============
0---50
1--184
2--286
3--356
4--394
5--400
6--374
7--316
8--226
9--104
10--has hit
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axis of symmetry is at vertex or t=4.6875
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negative values would imply the rocket was rising up through the building.
