## anonymous 5 years ago 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.

1. anonymous

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$984 = -\frac{1}{2}g(0)^2+c_1(0)+c_2 \rightarrow 984= c_2$

2. anonymous

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.