whats the answer?

Question

a ball is thrown upward from the top of a 25.0 m tall buiding the ball's initial speed is 12.0 m/s at the same instant, a person is running on the ground at a distance of 31.0 m from the building. what must be the average speed of the person if he is to catch the ball at the bottom of the building?

Answer 1

8.07 m/s

Answer 2

We need to know in this problem the time it takes the ball to reach the ground. Once we know that then we can compute for the average speed of the person running on the ground by using the formula, speed=distance/time.

Now let's follow the ball. If the initial speed u=12m/s, we can calculate the time it reaches the top of its flight using the formula, v=u+at. where v is the final speed(=0 at the top, right?), a the acceleration of gravity, 9.8m/s/s and t the time in sec. Thus

0=12+(-9.8)t . (g is negative because the ball is moving
opposite the force of gravity.)
t=-12/-9.8
=1.22 sec. This is the time it takes to reach the top of its flight. We multiply this by 2 to get the time it reaches the level of the top of the building, i.e. 2*1.22=2.44 sec.

Now we calculate the time it takes the ball to reach the ground, from a height of 25 meters. We use the formula s=ut +1/2at^2 where s=25meters, u=12m/s( you have to remember that the speed of the ball at the top of the building is the same going up as going down.)

25=12t+1/2(9.8)t^2
25=12t+4.9t^2
4.9t^2+12t-25=0

Solve for t using the formula:

t={-b+[(b^2-4ac)^1/2]}/2a
t={-b-[(b^2-4a3)^1/2]}/2a

Where a, b, and c are the coefficients of the 1st, 2nd,
and 3rd terms, respectively. (Please memorize this formula.) Now substitute the values of a,b,and c:

t={-12+[12^2-4(4.9)(-25)]^1/2}/2*4.9
=[-12+(144+490)^1/2]/9.8
=[-12+(634)^1/2]/9.8
=(-12+25.2)/9.8
=1.35 sec.

The other formula for t will give you a negative value, so we discard it, and use only t=1.35sec. We now add this
to 2.44sec which we obtained earlier, and we get the
time of 3.79sec.

Now substitute this in our formula speed=distance/time:

speed=31/3.79
=8.18m/s

Answer 3

read your physics text book. the answer to your question is there.