Answer:
a) v(t) = -32.2 ft/s² · t + 20 ft/s
b) h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²
c) The weight will reach its maximum height after 0.62 s. The maximum height will be 190 feet.
Explanation:
Hi there!
a) Since the only force that acts on the weight is the gravity force, the object is under a constant downward acceleration g = -32.2 ft/s² (it is negative because we consider the upward direction as positive). The acceleration is the variation of the velocity over time (dv/dt). Then:
dv/dt = g
Separating variables:
dv = g dt
Integrating from the initial velocity, v0, to v and from t = 0 to t, we obtain:
v - v0 = g t
v = g t + v0
Then:
v(t) = -32.2 ft/s² · t + 20 ft/s
b) The velocity of the weight is the variation of the height over time:
dh/dt = v(t)
dh/dt = g t + v0
Separating varibles:
dh = g t dt + v0 dt
Integrating from initial height, h0, to h and from t = 0 to t:
h - h0 = 1/2 · g · t² + v0 · t
h = h0 + v0 · t + 1/2 · g · t²
Then:
h(t) = 184 ft + 20 ft/s · t - 1/2 · 32.2 ft/s² · t²
h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²
c) When the weight reaches its maximum height, its velocity will be zero. Then, using the equation of velocity we can obtain the time at which the weight is at the maximum height:
v(t) = -32.2 ft/s² · t + 20 ft/s
0 = -32.2 ft/s² · t + 20 ft/s
-20 ft/s/ -32.2 ft/s² = t
t = 0.62 s
The weight will reach its maximum height after 0.62 s.
The maximum height will be h(0.62 s):
h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²
h(0.62 s) = 184 ft + 20 ft/s · (0.62 s) - 16.1 ft/s² · (0.62 s)²
h(0.62 s) = 190 ft
The maximum height will be 190 feet.
