A baseball player hits a ball. The ball leaves the bat with an initial upward velocity of 35 meters per second. If the ball is 1 meter off the ground when it leaves the bat, about how many seconds will it take for the ball to hit the ground?

I hate science and this seems to be related to that of an science question...could someone help?

2. This is a physics problem

Using conservation of energy...

$K_{i}+U_{i}=K_{f}+U_{f}$
$\frac{1}{2}mv_{i}^2+mgh_{o}=\frac{1}{2}mv_{f}^2+0$
$v_{i}^2+2gh_{o}=v_{f}^2$
$v_{f}=\sqrt{v_{i}^2+2gh_{o}}$

You could also use kinematics but this method is easier

3. Originally Posted by DrDank
This is a physics problem

Using conservation of energy...

$K_{i}+U_{i}=K_{f}+U_{f}$
$\frac{1}{2}mv_{i}^2+mgh_{o}=\frac{1}{2}mv_{f}^2+0$
$v_{i}^2+2gh_{o}=v_{f}^2$
$v_{f}=\sqrt{v_{i}^2+2gh_{o}}$

You could also use kinematics but this method is easier
Whoa lol confusing.
7.17 seconds-0.03 seconds 14.34 seconds 35.133 seconds
I think its 14.34 seconds, any thoughts?

if the maximum height is $h_{m}$
the initial height is $h_{o}=1 meter$
and the final height is $h_{f}=0 meters$
and g = -9.8 m/s

first find the time it takes to reach it's max height , $t_{1}$

$h_{m}-h_{o}=v_{o}t_{1}+\frac{1}{2}gt_{1}^2$ and $v_{m}^2=v_{o}^2+2g(h_{m}-h_{o})$

solve these for $h_{m}-h_{o}$ and set equal to eachother,

$v_{o}t_{1}+\frac{1}{2}gt_{1}^2=\frac{v_{m}^2-v_{o}^2}{2g}$

Now solve the quadratic for $t_{1}$, that's the time it takes to reach its max height.

That's only a portion of the time the total time is $t_{1}+t_{2}$, where $t_{2}$ is the falling time