MAY DAY MAY DAY, HELP


I have an urgent problem, 14 points to the answer-er!


Adam tossed a baseball into the air with an initial upward velocity of 10 meters per second.

AWrite an equation relating the distance in meters of the ball above where it was tossed t seconds after it was tossed

use dis d=rt - 5t^2

BHow high above where it was tossed would the ball be after 0.5

Verified Answer :

AL2006


Have you heard the expressions "shooting fish in a barrel" or
"taking candy from a baby" ?

I've decided to take your points and try not to feel guilty about it.

A).  That's it !  That's the equation, for ANY basketball on Earth. 
You wrote it right there in your question.

In that equation that you wrote, the ' r ' is the initial velocity. 
You said that Adam tossed it straight up, and it was going 10 meters per second
   when it left his hand. 
So ' r ' in the equation is +10.
The equation is:
                             d = 10t - 5t²    .
It tells you how high the ball is above its tossing height after any number of seconds.
The ' t ' in the equation is the number of seconds.  Any letter could have been used,
but ' t ' was cleverly selected because it stands for 'time'.

B).  You want to know where it is after 0.5 second ?    All you have to do is put '0.5'
into the equation wherever there's a ' t '.  Do you really need somebody to do that
for you ?

Well, OK.  You're being so overly generous with your points . . .

         Distance = 10 t - 5 t²

         Distance = 10 (0.5) - 5 (0.5)²

         Distance =     5       - 5(0.25)

         Distance =     5       -  1.25

         Distance =        3.75 meters

                                positive 3.75 means above Adam's hand. 

A thinking exercise:
-- He tossed the ball upwards at  10  meters per second.
-- Why is it not  5  meters above his hands after  0.5  second?