Question

A object is thrown down from the top of the Eiffel tower (984ft) with an initial velocity of 30 ft/sec. Write the equation of motion.

Answer 1

If the object is thrown straight down, the only component of motion will be in the vertical direction and the only acceleration acting upon it is that of gravity. We have,\[\frac{d^2y}{dt^2}=-g \rightarrow \frac{dy}{dt}=-g t+c_1 \rightarrow y=-\frac{1}{2}g t^2+c_1t+c_2\]When t is 0, you're at the top of the tower, so the position is y(0)=984ft; that is\[984 = -\frac{1}{2}g(0)^2+c_1(0)+c_2 \rightarrow 984= c_2\]

Answer 2

Looking the the first derivative (speed), we have at t=0, y'(0)=-30 (minus 30 because we've derived this equation in a coordinate system where the vertical axis increases with height...since we're throwing in the other direction, the velocity will be in the opposite direction).

So, y'(0)= -g(0) + c_1 ---> -30 = c_1

The equation of motion is therefore,\[y(t)=-16t^2-30t+984\]where I've taken the magnitude of acceleration to be 32ft/s^2.