Question

The height in meters of a projectile can be modeled by h = −4.9t2 + vt + s where t is the time (in seconds) the object has been in the air, v is the initial velocity (in meters per seconds) and s is the initial height (in meters). A soccer ball is kicked upward from the ground and flies through the air with an initial vertical velocity of 4.9 meters per second. After how many seconds does it land approximately?

Answer 1

we can think of this as just a normal negative quadratic equation. We need to find at what time (t) will the height (h) equal 0.

so if you plug in 4.9 for v, we get the quadratic equation 0 = -4.9t^2 + 4.9t
s = 0 since it starts on the ground

so now we have 0 = -4.9t^2 +4.9t

we can factor out a t from the left side, giving us:
0 = t(-4.9t +4.9)

so now we need to look at this and realize what possible values for t will satisfy the equation. Looking at this, we can tell that if t=0, or if t=1, than the equation is satisfied.

2 answers for t because the ball is at a height of 0 twice.. once right before we kick it, and once when it lands.

we know that t cannot equal 0, since that is the time when we kicked it

so the answer is t = 1

Answer 2

from the equation 0 = t(-4.9t + 4.9) to satisfy, we need to find when t(-4.9t+4.9) equals 0.

if t = 1, then 1*(-4.9(1) + 4.9) == -4.9 + 4.9... which is 0

Answer 3

well you said approximately, right?

1 is pretty close to .9574

Answer 4

well there are two ways to do this problem, using physics and using algebra. since your question said 'approximately', then i assumed this wasn't a physics problem

Answer 5

the way i answered is how to solve it using algebra

Answer 6

Okay! Thanks a bunch!

Answer 7

yw