Question

a ball is tossed up vertically and returns to earth 1 s later. what is the initial velocity and how high did it go?

Answer 1

\[h(t)=-16t^2+v_0t\] and i guess we know that
\[h(1)=0\] so we have
\[-16+v_0=0\]
\[v_0=16\] initial velocity is 16 feet per second ( if we are working in feet)

Answer 2

Yes, I believe 16 ft/s is correct, and the ball reached a max height of 4 ft.

Answer 3

its in meters

Answer 4

i cant figure out the equation for this to work

Answer 5

4.9 m/s initial velocity, 1.22 m max height

Answer 6

I'll help with the equation, hang on.

Answer 7

1.22 isnt right

Answer 8

The equation would be x(t) = x(0) + v(0)t + (1/2)at^2.
a is -9.8 m/s^2.
x(0) is assumed to be 0 if we are tossing from the ground.
Good so far?

Answer 9

So x(t) = v(0)t-4.9t^2.
We are interested in x(1)=0 at t=1 s, so ...

Answer 10

0 = v(0) (1) - 4.9 (1)^2
0 = v(0) -4.9
v(0) = 4.9 (m/s).

Answer 11

The max height is reached at t=1/2:
x(1/2) = 4.9(1/2) -4.9(1/2)^2, so ...

Answer 12

x(1/2) = 4.9 (1/4) = 1.225 (meters).

Answer 13

got it. its confusing.
why did you use 4.9?

Answer 14

4.9 = (1/2) (9.8).

Answer 15

You think it's supposed to be something other than 1.225 meters?

Answer 16

no its right.

Answer 17

i put 1.22