Help! throws a baseball vertically upward. s(t)=-16t^2+64t+192

[Xsan925]

A person standing close to the edge on the top of 192-foot building throws a baseball vertically upward.
s(t)=-16t^2+64t+192
How do I solve this?
a) After how many seconds does the ball reach the height of 112 feet?


b) How many seconds does it take until the ball finally hits the ground?

 

[Xsan925]

Need help.

A person standing close to the edge on the top of a 192-foot building throws a baseball vertically upward. The quadratic function: s(t)=-16t^2+64t+192

a) After how many seconds does the ball reach the height of 112 feet?


b)How many seconds does it take until the ball finally hits the ground?

also need help with number 4.

 

[Deleted member 4993]

A person standing close to the edge on the top of 192-foot building throws a baseball vertically upward.
s(t)=-16t^2+64t+192
How do I solve this?
a) After how many seconds does the ball reach the height of 112 feet?


b) How many seconds does it take until the ball finally hits the ground?

What are your thoughts regarding the assignment?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

 

[Dr.Peterson]

A person standing close to the edge on the top of 192-foot building throws a baseball vertically upward.
s(t)=-16t^2+64t+192
How do I solve this?
a) After how many seconds does the ball reach the height of 112 feet?


b) How many seconds does it take until the ball finally hits the ground?

Which variable (s or t) corresponds to the 112 feet?

Which variable gives the number of seconds that you are asked to find?

How can you find that value?

Then, which variable do you know when the ball hits the ground?

 

[HallsofIvy]

If we were to interpret this strictly, "A person standing close to the edge on the top of 192-foot building throws a baseball vertically upward" then the ball will never get down to 112 feet because it hits the top of the building! But if the ball is thrown not quite vertically so that it misses the top of the building on the way down, then we can do this.

"s(t)=-16t^2+64t+192" So the baseball was thrown up at a speed of 64 ft/s. The 192 ft is the height of the building, of course, and the -16 is the acceleration due to gravity.

"a) After how many seconds does the ball reach the height of 112 feet?"
s(t) in the equation above is the height above the ground. Solve the quadratic equation -16t^2+ 64t+ 192= 112.

"b)How many seconds does it take until the ball finally hits the ground?"
"The ground" is at height s(t)= 0. Solve the quadratic equation -16t^2+ 64t+ 192= 0.

(It helps that 64, 192, and 112 are divisible by 16.)

 

[Dr.Peterson]

If we were to interpret this strictly, "A person standing close to the edge on the top of 192-foot building throws a baseball vertically upward" then the ball will never get down to 112 feet because it hits the top of the building! But if the ball is thrown not quite vertically so that it misses the top of the building on the way down, then we can do this.

I think he is reaching out past the edge, so that his hand is not above the building, and he can in fact throw the ball vertically.

What I'm not sure of is the relationship of the height of his hand to the height of the building.