Posted by **Anonymous** on Wednesday, February 2, 2011 at 12:32pm.

A ball is shot straight up into the air with initial velocity of 47 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

- Calculus -
**helper**, Wednesday, February 2, 2011 at 2:11pm
A ball is shot straight up into the air with initial velocity of 47 ft/sec

Free fall, s in ft, t in sec

s = s0 + v0t - 16t^2

height, s0 = 0

velocity, v0 = 47

s = 0 + 47t - 16t^2

s = 47t - 16t^2

Instantaneous velocity, v

v = ds/dt = v0 - 32t

v = 47 ft/sec

v = 47 - 32t

To find time, t

At maximum value or turning point, v = 0

v = 47 - 32t

0 = 47 - 32t

32t = 47

t = 47/32 = 1.47 sec to reach max height

s = 47t - 16t^2

t = 1.47 sec

s = 47(1.47) - 16(1.47)^2

s = 69.09 - 16(2.16)

s = 69.09 - 34.56

s = 34.53 feet

So, it takes 1.47 sec to reach its maximum height of 34.53 feet

Check my math

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