Calculate (1) how high it goes, (2) how long the ball is in the air, (3) how much time it takes for the ball to reach the maximum height,

Question

A person throws a ball upward into the air with an initial velocity of 15 m/s with

an acceleration of -9.8m/s². Calculate (1) how high it goes, (2) how long the

ball is in the air, (3) how much time it takes for the ball to reach the maximum

height, (4) the velocity of the ball when it returns to the thrower’s hand.

Include:

Important Notations

Useful Given

Solution:

Velocity Function

Position Function

Final answer in (1)

Final answer in (2)

Final answer in (3)

Final answer in (4)

Interpretation

Answer 1

h(t) = -4.9t^2 + 15t + ho where t = time in seconds, h=height in meters, ho = initial height

h'(t) = velocity = v(t) = -9.8t +15

set velocity = 0 to find maximum height

velocity = the derivative of the height function

-9.8t = -15

t = 15/9.8 = about 1.531 seconds

substitute t = 15/9.8 into the height function to find the maximum height

h(15/9.8) = maximum height = -4.9(15/9.8)^2 +15(15/9.8) + ho

h(15/9.8) = (15/2)(15/9.8) +ho

h(15/9.8) = 225/19.6 + ho

h(15/9.8) = about 11.48 + ho meters high

the ball's velocity when it returns to the thrower's hand is the same speed as when it left the thrower's hand, 15 meters per second, only in the opposite direction. vo=15 m/sec. On return, it's velocity = -vo = -15 m/sec

the ball was in the air twice as long as it took to reach maximum height

2 x 15/9.8

= 30/9.8 seconds

= about 3.09 seconds

ho = height of the thrower, maybe a little less than 2 meters