Here you will derive the position function s(t) for an object tossed into the ai
ID: 2861770 • Letter: H
Question
Here you will derive the position function s(t) for an object tossed into the air. Recall that the velocity function is v(t) = s'(t) and the acceleration function is a(t) = s"(t). Assume "up" is the positive direction. Acceleration is due to gravity, g (e.g., -9.8 m/s^2 or -16 ft/s^2). The initial velocity of the object is written v(0) = v_0 and the height from which the object is tossed is s(0) = h. To get the position function, solve the initial value problem: Find s(t), given s"(t) = g, s'(0) = v_0, and.s(0) = h. If we include air resistance, the formula gets more complicated. The velocity function in this case is v(t) = -mg/beta + (mg/beta + v_0) e^-beta/m t, where m is the mass of the object and beta is a constant called the drag coefficient. Using s(t) = integrate v(t) dt and s(0) = h, rewrite the position function with air resistance.Explanation / Answer
(a) s''(t) = g
=> s'(t) = gt + c
=> s'(0) = c = v0
=> s'(t) = gt + v0
Integrating the above equation
=> s(t) = 0.5gt2 + v0t + k
=> s(0) = k = h
=> s(t) = 0.5gt2 + v0t + h
(b) Integrating the equation we get
s(t) = -(mg/)t + (-m/)( (mg/) + v0 )e(-/m)t + k
s(0) (-m/)( (mg/) + v0 ) + k = h
=> k = h + (m/)( (mg/) + v0 )
=> s(t) = -(mg/)t + (m/)( (mg/) + v0 )( 1 - e(-/m)t ) + h