Baseball Star Bryan is standing at the top of the Sears Tower in Chicago and decides to throw a baseball up with a velocity of 5 m/s. The Sears Tower is 442 meters tall and gravity exerts a constant acceleration of -9.8 m/s/s. Ignore ball mass and wind resistance.

Question

a.) Find equations for acceleration, velocity, and position.
b.) At what time does the ball reach its highest point? How high is that point?
c.) When does the ball hit the ground? How fast is the ball traveling when it hits the ground?

Answer 1

a = -9.8
then v = -9.8t + c
when t = 0 , v = 5 m/s
5 = 0 + c, ----> c = 5
v = -9.8t + 5

s = -4.9t^2 + 5t + k
when t = 0, s =442
442 = 0 + 0 + k

s = -4.8t^2 + 5t + 442

b) highest point when v = 0
0 = -9.8t + 5
t = 5/9.8 = appr .51 sec

when t = .51
s = -4.9(.51)^2 + 5(.51) + 442 = 443.28 m

c) when it hits the ground, s = 0
-4.9t^2 + 5t + 442 = 0

use the quadratic formula to solve this equation
One answer will be negative, of course you will have to reject that one.
sub the positive value of t into your velocity equation.

Answer 2

a = -g = -9.8 m/s^2 The statement gives you the answer.

v = integral a dt = -9.8 t + c
at t = 0
v = Vi = 5 m/s
so c = 5
v = -9.8 t + 5

h = integral v dt = 5 t -(9.8)/2 t^2 + cc
when t = 0
h = Hi = 442
so cc = 442
h = 442 + 5 t - 4.9 t^2 (check your physics book :)

at top, v = 0
0 = 5 - 9.8 t
solve for t

0 = 442 + 5 t - 4.9 t^2
solve quadratic for t when h = 0 the ground

then as usual v = 5 - 9.8 t