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.
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
