Answer:
43 + (8.93)*t + (9.81)t²/2 for 0 < t < 0.91
55 - (9.81)t²/2 for 0.91 < t < 3.06 sec
Step-by-step explanation:
Simple kinematic equations used:
a = dv/dt ... Eq 1 and v = ds/dt .... Eq 2
Using Eq 1:
a.dt = dv - separating variables
∫a.dt = ∫dv - integrating both sides 0< t < t and vi < v < vf
(a.t) = (vf - vi) - Solving and evaluating integral
Hence,
vf = vi + a*t .... Eq 3
Using Eq 2:
vf.dt = ds - separating variables
∫(vi+a*t).dt = ∫ds - Substitute Eq 3 and integrating both sides 0< t < t and si < s < sf
vi*t + at²/2 = sf-si .... Eq 4
Now using the data given and Eq 4 & Eq 3:
Interval 1: For rock at max height above road @ t = 0.91 sec and vf = 0, for 0 < t < 0.91
vi = - a*t = - (-9.81)*(0.91) = 8.93 m/s
Plugging in Eq 4:
sf = si + vi*t + at²/2 = 43 + (8.93)*t + (9.81)t²/2 = 55 fts @ t = 0.91 sec
Interval 2: For rock in contact with road @ t = 3.06 sec and vi = 0, for 0.91 < t < 3.06
Using Eq 4
sf = si + vi*t + at²/2 = 55 - (9.81)t²/2 for 0.91 < t < 3.06 sec
