Time to reach Max. Height for a kicked soccer ball

AI Thread Summary
The discussion centers on calculating the time it takes for a kicked soccer ball to reach its maximum height, with the initial vertical velocity calculated as 7.5 m/s. Participants highlight that the original approach incorrectly sets the distance (d) to zero instead of focusing directly on time (t). The correct method involves recognizing that at the highest point, the vertical velocity is zero, and using the formula t = v/g to find the time to reach maximum height. The final consensus confirms that the time to reach maximum height is indeed 0.76 seconds, but emphasizes the need for clarity in the problem-solving process. Understanding the physics of upward motion under gravity is essential for accurate calculations.
physicsstudent111
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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?
 
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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?
 
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what is the vertical velocity at the balls highest point?
 
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The Vertical velocity is 7.5m/s
 
physicsstudent111 said:
The Vertical velocity is 7.5m/s
At the highest point? nope
 
physicsstudent111 said:
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.
olivermsun said:
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.
 
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physicsstudent111 said:
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}##
 
olivermsun said:
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.
 
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jbriggs444 said:
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.
 
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