An object moves from O, at t = 0, to A at the edge of the surface of a cliff whi
ID: 1487627 • Letter: A
Question
An object moves from O, at t = 0, to A at the edge of the surface of a cliff which is 6.2 m above the ground as shown in the figure. Between O and A, the position of the object is observed to vary with time as x(t) = o. 4012 + 0. 50t, where x is in meters and t is in seconds. It observed that the object reaches point A at t = 5.0 s. Determine the instantaneous horizontal velocity at point A. The object leaves the surface at A with the horizontal velocity obtained in part (a) and follows the path shown before striking the ground at D. Determine the time T taken by the object to strip e e ground after leaving A. Determine the distance BD between the bottom edge B of the cliff and the point D where the object strikes the ground. Determine me magnitude of the velocity of the object just before K reaches the ground.Explanation / Answer
a)
x=0.40t^2+0.50t ----------(1)
v=dx/dt = 0.80t + 0.50 -----------(2)
a=dv/dt= 0.80 m/s^2 -----------(3)
Plugging t=5s in (2)
v(5) = 0.80t + 0.50 = 0.80*5 +0.50 = 4.5m/s
Thus vix = 4.5m/s
b) Use equation
y= viy*t + 1/2*a(y a)*t^2
Since object is moving horizontally, viy = 0 m/s
a(y) = g = -9.8 m/s^2
Plugging values,
-6.2 = 0*t -1/2*9.8*t^2 => t= 1.12s
c) d(BD) = vix*t = 4.5*1.12 = 5.04 m
d) vfx = v(5+1.12) = 0.80(5+1.12) + 0.50 = 5.4 m/s
Vfy = viy -gt = 0 -9.8*1.12 = -11.0 m/s
vf = sqrt(vfx^2 + vfy^2) = sqrt(5.4^2 + 11.0^2) = 12.25 m/s