Question 964576: A tennis ball is launched upward at 32 feet per second from a platform that is 15 feet tall.
Find the maximum height the ball reaches using the equation that describes the height of the tennis ball, based on time, initial velocity, and initial height.
h(t)=−16t2+96t+200
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A tennis ball is launched upward at 32 feet per second from a platform that is 15 feet tall.
Find the maximum height the ball reaches using the equation that describes the height of the tennis ball, based on time, initial velocity, and initial height.
h(t)=-16t^2+96t+200
complete the square:
h(t)=-16(t^2-6t+9)+144+200
h(t)=-16(t-3)^2+344
The ball reaches the maximum height of 344 ft 3 seconds after it is launched
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
A tennis ball is launched upward at 32 feet per second from a platform that is 15 feet tall.
Find the maximum height the ball reaches using the equation that describes the height of the tennis ball,
based on time, initial velocity, and initial height.
h(t)=−16t2+96t+200
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In the post, the given equation does not correspond (contradicts) to the wording part.
So, the best thing that a reader can do with this post is to throw it to the closest garbage bin.
Do not accept the "solution" by @lwsshak3, since it is irrelevant to the problem and uses
totally different input data.
|
|
|
