Calculus
posted by Anonymous .
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?

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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