Question

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?

Answer 1

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

Answer 2

??

Answer 3

equation is \[h(t)=25t-16t^2\]

Answer 4

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

Answer 5

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

Answer 6

then plug it in

Answer 7

I hope my equation and calculation is correct.

Answer 8

maybe it is let me check

Answer 9

thank youuu guys :)

Answer 10

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

Answer 11

m/s^2

Answer 12

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

Answer 13

\[\frac{1775}{64}=27.73\]

Answer 14

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

Answer 15

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

Answer 16

using 32 ft/s^2 for gravity