A ball is dropped from the top of the Empire State building to the ground below. The height, y, of the ball above the ground (in feet) is given as a function of time, t, (in seconds) by

y=1250-16t^2

(a) Find the velocity of the ball at time t. What is the sign of the velocity? Why is this to be expected?

(b) Show that the acceleration of the ball is a constant. What are the value and sign of this constant?

(c) When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/ sec = 15/22 mph).

Question

A ball is dropped from the top of the Empire State building to the ground below. The height, y, of the ball above the ground (in feet) is given as a function of time, t, (in seconds) by
 
 	y=1250-16t^2
 
 	(a)  	Find the velocity of the ball at time t. What is the sign of the velocity? Why is this to be expected?
 
(b)  	Show that the acceleration of the ball is a constant. What are the value and sign of this constant?
 
(c)  	When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/ sec = 15/22 mph).

anonymous · Answer

(a) Use one of the kinematic equations for constant acceleration: $$v = v_{0} + at$$You know the acceleration (-9.81m/s^2) and the initial velocity (0), so a function for the velocity at a given time t is:$$v = 0 + -9.81\frac{m}{s^{2}}t = -9.81\frac{m}{s^{2}}t$$
The sign of the velocity is negative because it's falling, not going up.