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Motion problems (Calculus)?
1. A rock thrown vertically reaches a height of s(t) = 24t - 4.9t^2
a) How high will the rock go?
b) How long would it take to reach half its max. height?
c) How long would the rock be aloft?
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2. A 45-caliber bullet is fired straight up from ground level with velocity 832 ft/sec. How long would the bullet be aloft and how high would it go?
Thanks! :] I have my math final today. T-T
2 Answers
- TLv 59 years agoFavorite Answer
When the rock reaches its max height, the velocity will be 0. Derivative of speed = velocity.
v(t) = 24 - 9.8t
0 = 24-9.8t
t = 2.45 s
1a) s(2.45) = 24(2.45) - 4.9(2.45)^2 = 29.39 height units
1b) 14.7 = 24t - 4.9t^2 4.9t^2 - 24t + 14.7 = 0 Use quadratic t = 0.72 s on the way up 4.18 s on the way down.
1c) 0 = 24t - 4.9t^2 t=0(start) and t = 4.9 (end) or 2(time to reach peak) = 2(2.45) = 4.9
2) solve the same way with s(t) = 832t - 32.2t^2
- grunfeldLv 79 years ago
a) s ' ( t ) = 24 - 9.8t
0 = 24 - 9.8t
9.8t = 24
t = 24 / 9.8
s( t ) = 24*24/9.8 - 4.9(24 / 9.8)^2 = ??
b) ??( from a ) / 2 = 24t - 4.9t^2
solve the equation.
c) 0 = 24t - 4.9t^2
0 = t(24 - 4.9t)
t = 0, (24 / 4.9)
(24 / 4.9) - 0 = 24 / 4.9
2. you can work the question.
Source(s): my brain
