Question

The height of a falling object is given by h(t) = 400 − 39t − 157e−t/4 with h in meters and t in seconds. Assuming that the ground is at height h = 0, how fast is the object moving at the instant it hits the ground? Your answer must be accurate to one decimal place.

Answer 1

This is the concept of algebra; given that the height of the house at time t is given by:
h(t)= 400 − 39t − 157e−t/4
the velocity of the house is given by:
dh/dt
velocity of our object is:
dh/dt=-39+1/4*157e^-(t/4)
=-39+157/4 e^(-t/4)
thus the velocity at h=0 will be as follows:
at h=0, the value of t will be:
0=400-39t-157e^(-t/4)
solving for t we get:
t=-5.4 or t=10.1
since we don't have negative time, t=10.1
thus the velocity at this point will be:
dh/dt=-38+1/4*157e^(-10.1/4)
=-38.86 sec
this means that the object dropped by a velocity of 38.86 sec