Word Problem: Find height of ball after 1 second

[pooh27]

Suppose a baseball is shot up from the ground straight up with an initial 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=-16t^2+vot +so.

16 represents 1/2g, the gravitional pull due to gravity(measured in feet per second^2)

Vo is the intial velocity (how hard do you throw the object, measured in feet per second)
So is the initial distance above ground( in feet) If you are standing on the ground, then So=0

The questions I have is what function describes this problem and how high will the ball be after 1 second.

 

[skeeter]

Re: Word Problem

Suppose a baseball is shot up from the ground straight up with an initial 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=-16t^2+vot +so.

16 represents 1/2g, the gravitional pull due to gravity(measured in feet per second^2)

Vo is the intial velocity (how hard do you throw the object, measured in feet per second)
So is the initial distance above ground( in feet) If you are standing on the ground, then So=0


The questions I have is what function describes this problem and how high will the ball be after 1 second.

the function was given to you ...

s=-16t²+vot +so

substitute the given parameters into the equation ...

vo = 64
so = 0 (because it was "shot" up from ground level)
t = 1 sec

s = -16(1)² + 64(1) + 0

s = ?

 

[pooh27]

How long will it take to hit the ground?

What is the maximum heigth of the ball?

What time will the maximum height be attained?

 

[skeeter]

How long will it take to hit the ground?

set s = 0 and solve for t.

What time will the maximum height be attained?

in this case (since the ball was launched from ground level), at half the time calculated for the first question.

What is the maximum heigth of the ball?

calculate s for the time of max height.

 

[TchrWill]

Suppose a baseball is shot up from the ground straight up with an initial 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=-16t^2+vot +so.

16 represents 1/2g, the gravitional pull due to gravity(measured in feet per second^2), Vo is the intial velocity (how hard do you throw the object, measured in feet per second), So is the initial distance above ground( in feet) If you are standing on the ground, then So=0

The questions I have is what function describes this problem and how high will the ball be after 1 second.

You already have the expression that defines the height of the ball in h = Vot - gt^2/2.

)After 1 second, h = 64(1) - 32(1^2) makink h = 32 ft.

a)How long will it take to hit the ground?

b)What is the maximum heigth of the ball?

c)What time will the maximum height be attained

The time to reach its maximum height derives from Vf = Vo - gt or 0 = 64 - 32t making t = 2 seconds. Since it takes the same time to fall back to the ground, the total time from start to finish is 4 seconds.

The maximum height reached derives from h = 64(2) - 16(2^2) from which h = 64 ft.

The maximum height is reached in 2 seconds as calculated above