Question

Suppose you throw a baseball straight up at a velocity of 64 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
� 16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
� v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
� s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
1. How long will it take to hit the ground?
2. What is the maximum height of the ball? What time will the maximum height be attained?

Answer 1

Answer:

1. t = 4s

Step-by-step explanation:

Initial Velocity vo = 64

s = o when the ball is thrown from the ground or when the ball is on the ground.

1. For the time taken for the ball to reach the ground:

Substituting s = 0, vo =64 and so = 0 (since it was thrown from the ground)in the equation, we'll have

s= -16t2 + vot + so  we will get

0 = -16t2 + 64t

solving for t, we get the common factor or the right side of the equation

0= 16t(-t + 4)

Dividing both sides by 16t

0 = -t + 4

We need to find the value of t so we move it to one side of the equation

t= 4secs

2. If it takes 4secs from throwing to when it returns, it will take half the time to reach its highest point.

To find the highest possible height, we will substitute t=2 into the initial equation.

We'll have

s= -16*(2)2 + 64 *(2) +0

s= -16*4 + 128

s= -64 + 128

s= 64 ft

The maximum height it will attain is 64ft.