1. The height (in meters) of a projectile shot vertically upward from a point 2
ID: 2864770 • Letter: 1
Question
1. The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22.5 m/s is h = 2 + 22.5t 4.9t2 after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
v(2) = _____________m/s
v(4) = ______________m/s
(b) When does the projectile reach its maximum height?
___________________s
(c) What is the maximum height?
________________________m
(d) When does it hit the ground?
______________________________s
(e) With what velocity does it hit the ground?
__________________________________m/s
2. Differentiate the function.
y = ln|4 + t t3|
y' =
Explanation / Answer
1.
(a) Differentiate h(t) with respect to 't' to get v(t)...
h(t) = -4.9t^2 + 22.5t + 2
v(t) = -9.8t + 22.5
v(2) = -19.6 + 22.5 = 2.9 m/s
v(4) = -39.2 + 22.5 = -16.7 m/s, or 16.7 m/s downwards
(b) Height is reached when v(t) = 0, so...
-9.8t + 22.5 = 0
t = (22.5/9.8) = 2.2959 seconds
(c) Solve for height using this time...
h_max = 2 + 22.5(22.5/9.8) - 4.9(22.5/9.8)^2
h_max = 27.829 meters
(d) Let h(t) = 0...
h(t) = -4.9t^2 + 22.5t + 2 = 0
Use the quadratic formula...
t = [-22.5 ± sqrt(22.5^2 - 4(-4.9)(2))]/-9.8
t = [-22.5 ± sqrt(545.45)]/-9.8
t = [-22.5 ± 22.35]/-9.8
t = -0.0872, t = 4.5765
So t = 4.5765 seconds.
(e) Find v(t) using this time...
v(4.5765) = -9.8(4.5765) + 22.5
v(4.5765) = -22.35 m/s, or 22.35 m/s downwards
2. y = ln l 4 + t - t3l
y' = dy/dt = 1/[ l 4 + t - t3l * d/dt( l 4 + t - t3l )
= l 1 - 3t2l / l 4 + t - t3l
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