Finding Time for a Kicked Ball

[balancedlamp]

Problem: You are playing soccer with some friends while some people nearby are playing hide and seek. When the seeker gets to '4', you kick the soccer ball high in the air. WHen the seeker says '5', the ball is 10 ft in the air. When the seeker gets to '7', the ball is 20 feet in the air. How many seconds will the ball be in the air in total before it hits the ground?

Given that after 1 second the ball traveled 10 ft in the air, I had the initial velocity as 10 ft/s. At the apex, the velocity will be 0 ft/s. Since change in position is not given I used the equation Vf = Vi + at. My thought was to find the time this way then multiply by two. Using a=-32.2 ft/s/s, I got t=5/8 s at the end of it all, which doesn't make any sense. I acknowledge that I left out the part about it being 20ft in the air after 2 seconds, but I thought that would erroneous considering we know the initial velocity and the value of a. What am I doing wrong?

 

[skeeter]

In my opinion,the problem is either bogus or the setting of the problem isn't on Earth's surface.

You are given 3 data points (t,h), where t is time in seconds and h is in ft ...

(0,0) , (1,10) , (3,20)

those three data points yield the quadratic $h = \dfrac{35}{3} \cdot t - \dfrac{5}{3} \cdot t^2$

if done on Earth,the equation should be $h = v_0 \cdot t - 16t^2$

 

[I like Serena]

Perhaps the 'seeker' calls the time every 3 seconds or so?

 

[balancedlamp]

Once I approached it using the three points and finding the quadratic equation, I just plugged it into a graphing calculator and got the answer of 7 seconds, which is one of the answer choices.

However, I'm confused why the kinematic equations are giving me such funky answers, while the quadratic equation is pretty straightforward. Was my approach using the initial kinematic equation wrong?

 

[HallsofIvy]

Once I approached it using the three points and finding the quadratic equation, I just plugged it into a graphing calculator and got the answer of 7 seconds, which is one of the answer choices.

However, I'm confused why the kinematic equations are giving me such funky answers, while the quadratic equation is pretty straightforward. Was my approach using the initial kinematic equation wrong?

You say you used "

a=-32.2 ft/s/s" but, as skeeter said, that, the Earth's acceleration due to gravity, does not give the data in the problem. With the given data, the acceleration due to gravity, wherever this is, is $$\frac{10}{3}$$ feet per second per second, not 32.2 feet per second per second.

Or, as Klaas van Aarsen points out, the problem does not actually say that the person calling numbers calls one per second. Taking the trajectory to be $$(35/3)x- (5/3)x^2$$ then $$\frac{5}{3}x^2= 16.1 t^2$$ if $$x^2= \frac{48.3}{5}t^2$$ so that $$x= \sqrt{\frac{48.3}{5}}t= \sqrt{9.66}t= 1.932 t$$ so that the person was calling numbers every 1.932 seconds or approximately every 2 seconds, not every three seconds.

Of course, that would change the answer. If the ball was in the air for x= 7 "units" then it is in the are for [tex]7(1/932)= 13.524[/tex] seconds, approximately 14 seconds.
​

 

[balancedlamp]

My apologies, according to the problem, the seeker counts at a rate of 1 whole number per second. Sorry for leaving that out, that was completely accidental.