Why is a ball thrown in the air symmetrical in the time it falls I’m having trouble with high school physics question.
The question is finding the time in the air of a ball that is thrown into the air with some initial velocity (negligible air resistance) and reaches its maximum height at $0~\rm m/s$ and gravity is $9.81~\rm m/s^2$.
I found the time it takes to reach the maximum height using $v_f = v_i + at$. However, my answer book says that the time to fall is symmetrical to the time it takes to reach maximum height.
but isn’t the time it takes to fall from a maximum height calculated with $$h= \frac{1}{2} a t^2$$ How could this be symmetrical with the initial velocity of $5~\rm m/s$ if it falls from $0~\rm m/s$?
 A: When going up the velocity of the ball starts at 5 m/s and ends at 0m/s. When going down the process is reversed: starts at 0m/s and ends at 5m/s. The change in velocity is symmetrical.
Since the acceleration is the same in both cases, the time it takes to make this velocity change is the same.
For further clarification: your formula to find the time it takes to reach the maximum height also works for the time it takes for the ball to fall from the maximum height, but this time $v_0$ is 0 m/s and $v_f$ is -5 m/s.
A: So the three equations of motion we have are:
$$v = u + at$$
$$s = ut + \frac{1}{2}at^2$$
$$v^2 - u^2 = 2as$$
where:

*

*u is the initial velocity

*v is the final velocity

*s is the total displacement

*t is the total time taken

*a is the constant acceleration

for a body thrown up in Earth's gravity and caught at the same point as the point of throwing will have the following values:
$$v = -u$$
$$s = 0$$
$$a = -9.8 m/s^2$$
simply plugging these values into the first equation we get:
$$t = \frac{2u}{9.8}s$$
this answers the question asked. Now moving on to your concerns, we have two cases the upward motion and the downward motion. they have the following values:
$$(I)$$
$$u = 5m/s$$
$$v = 0 m/s$$
$$a = -9.8m/s^2$$
$$(II)$$
$$u = 0m/s$$
$$s(II) = -s(I)$$
$$a = -9.8m/s^2$$
Plug these values in the three equations of motion to get your values for t, v(II) and s which will show a clear symmetry.
A: It is easiest to understand by realizing that once it leaves the hand it is under the same amount of gravitational acceleration going up as it is coming down. Only the signs + and -  change for accelerations. So the time to go up is the same as the time to go down.
A: You know how height and velocity of the ball depend on time:
$$\begin{align}
h(t)&=v_0t-\frac{1}{2}gt^2 \\
v(t)&=v_0-gt
\end{align} \tag{1}$$
where $g$ is the gravity acceleration and $v_0$ is the initial velocity.
You also wrote you know how to calculate time of maximal height,
which is $t_m=\frac{v_0}{g}$.
I suggest you do the following exercise.
Rewrite the above equations (1) using $t-\frac{v_0}{g}$
instead of $t$. After a few lines of algebra you will get the result:
$$\begin{align}
h(t)&=\frac{v_0^2}{g}-\frac{1}{2}g\left(t-\frac{v_0}{g}\right)^2 \\
v(t)&=-g\left(t-\frac{v_0}{g}\right)
\end{align} \tag{2}$$
From this result you can directly see that the movement
is symmetrical to time $\frac{v_0}{g}$.
