## anonymous 5 years ago Jules kicks a soccer ball off the ground and into the air with an initial velocity of 25 feet per second. Assume the starting height of the ball is 0 feet. Approximately what maximum height does the soccer ball reach?

1. anonymous

By Newton's third law of motion, v^2=u^2+2as where, "v" denotes final velocity, "u" denotes initial velocity, "a" denotes acceleration, and "s" denotes the height attained. Here, "v"=0. Therefore, (-u)^2=2as Or, -25*-25=2*10*s Or, 625=20s Or, s=625/20=31.25 feet

2. anonymous

??

3. anonymous

equation is $h(t)=25t-16t^2$

4. anonymous

maximum is given by $\frac{-b}{2a}$ and here $a=-16,b=25$ get $\frac{-b}{2a}=\frac{25}{32}$

5. anonymous

there shouldnt be anything wrong with using v^2 -u^2 +2as

6. anonymous

then plug it in

7. anonymous

I hope my equation and calculation is correct.

8. anonymous

maybe it is let me check

9. anonymous

thank youuu guys :)

10. anonymous

aadarsh the velocity is feet per second and you used m/s for gravity thats the only problem I think

11. anonymous

m/s^2

12. anonymous

i get \frac{1775}{64}=27.73 \]

13. anonymous

$\frac{1775}{64}=27.73$

14. anonymous

ya, ya!! ur correct. Let me do the correction. Or, -25*-25=2*10*10/3*s Or, 625=200/3s( Since, a=10 m/s^2, now, 1m=100 cm, and 30 cm=1 foot, so, 1 m=10/3 feet) Or, 625*3=200s Or, 1875=200s Or, s=1875/200=9.375 feet

15. anonymous

I did it both ways and got 625/64=9.77 feet both times

16. anonymous

using 32 ft/s^2 for gravity