# Time to reach Max. Height for a kicked soccer ball

• physicsstudent111
Of course, you have to convert the initial vertical velocity to a vertical velocity at the highest point, which is what you did.In summary, the ball reaches its highest point when it stops moving vertically, which occurs when the vertical velocity is zero. Using the SUVAT equation ##v=at##, we can find the time it takes for the vertical velocity to reach zero, which is the time when the ball reaches its highest point.f

#### physicsstudent111

Homework Statement
If a soccer player kicks a ball from the ground with an initial velocity of 15 m/s [30 degrees] above the horizontal] how long does it take the ball to reach its maximum height?
Relevant Equations
Vvertical=dsintheta
d=Vvertical*t+0.5atsquared
Vvertical=15sin30
= 7.5m/s

d=Vvertical*t+0.5atsquared
0=(7.5m/s)t+0.5(-9.81m/s)t squared
0=t(7.5m/s-4.9t)
t= 1.53s

t=1.53/2
t= 0.76s to reach the maximum height

Is this correct?

The answer is correct, but the steps you show have some issues.
Why are you solving for d in the first place? The question asks for t, not d.
In the second step, why are you setting d = 0?

topsquark
what is the vertical velocity at the balls highest point?

topsquark
The Vertical velocity is 7.5m/s

The Vertical velocity is 7.5m/s
At the highest point? nope

The Vertical velocity is 7.5m/s
No, that’s the initial vertical velocity. @malawi_glenn asked for the vertical velocity at the highest point.

SUVAT equations, in one standard form, involve five variables:
S distance
U initial velocity
V final velocity
A acceleration (which must be constant)
T time elapsed
Correspondingly, there are five equations, each omitting one variable.
The usual way to use them is to identify which variable you don't care about and pick the equation that omits it. In the present case you know u, v and a, and wish to find t. Which equation does not involve s?

That said, your method here, finding the time to return to the ground and halving it, is fine.
Why are you solving for d in the first place? The question asks for t, not d.
In the second step, why are you setting d = 0?
See above.

malawi_glenn
Vvertical=15sin30
= 7.5m/s

t= 0.76s to reach the maximum height

Is this correct?
note that ##0.76s = \dfrac{7.5m/s}{g}##

The answer is correct, but the steps you show have some issues.
Why are you solving for d in the first place? The question asks for t, not d.
In the second step, why are you setting d = 0?
As in the work of many students, no description of the strategy is provided. We are left to guess at the motivation for writing down various equations. Or the justification. We are forced to reverse engineer.

We set d=0 because that is the altitude of the soccer ball once it lands on the ground again.

So we solve that quadratic. It's an easy quadratic because the constant term is zero, so we can skip the completing the square or the quadratic formula. We get the time until the ball is back on the ground. We conveniently ignore the other solution at t=0.

Then we divide by two because the high point (and the time for the high point) is midway between the end points.

PeroK
As in the work of many students, no description of the strategy is provided. We are left to guess at the motivation for writing down various equations. Or the justification. We are forced to reverse engineer...

I asked because thought it might be instructive for the OP to walk us through his or her thinking.

From a mathematics perspective the solution is not incorrect. You have an unknown and a collection of equations to choose from. If you can find one that relates the unknown to the knowns and you can solve for it then it's fine.

From a physics learning standpoint I think it's useful to focus on the original question: When does the ball reach its highest point? Put another way, What does it mean for an upward-traveling object to reach its highest point under constant (downward) acceleration by gravity? malawi_glenn alluded to this in a few earlier posts.

Once you realize that gravity will decelerate the object until it's stops moving (vertically) at the top of the trajectory, then all you need is the definition of acceleration ##v=at##, and you have $$t_{v=0} = \frac{v}{-g},$$ which is what PeroK pointed out.