# Math Question For Algebra

User Generated

qq123

Mathematics

### Question Description

If a ball is thrown upward at 64 ft per second from a height of 12 ft, the height of the ball can be modeled by S=12=64t-16^2, where t is the number of seconds after the ball is thrown. how long after the ball is thrown is the height 40 ft?

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v1= 64 feet per second [up]
a= 9.81 m/s2 [down]= -9.81 m/s2 [up] = -32.185 feet/s2 [down] (1 metre= 3.28083989501312 feet)
t= 3s
d= ?

Step 2
Solve:

d= v1t+(1/2)at2
d= 64(3)+(1/2)(-32.185)(3)2
d= 192-144.8325
d= 47.165 feet

*Remember you need to add the initial height

d=12+47.165
d=59.1675 feet= 10 feet

rhtrarzvxr (18)
UC Berkeley
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