A baseball is hit at 45 m/s at 45degrees above the horizontal. How long did it remain in the air?

Answer 1

Make an assumption: no air resistance.

so the only thing we care about is the vertical displacement so we need to find the vertical velocity.

this is simply

45 m/s * sin(45) = 31.82 m/s

when it comes back down, the velocity will have the same magnitude but different sign at h = 0.

Thus

v_f = v_0 + at

a = -9.81 (acceleration of gravity)

-31.82 = 31.82 - 9.81 * t

t = 6.49 s

Answer 2

At its highest point, the velocity is only in the horizontal direction, so you can set the vertical velocity equal to zero.

v(t) = v + at
0 = (45 m/s)(sin 45°) + (-9.81 m/s²)t
0 = (31.8198 m/s) + (-9.81 m/s²)t
t = 3.2436 s

This is the halfway point, so the total time before hitting the ground is simply 2t, or about 6.4872 s.

Answer 3

Depending on Photon Bombardment and wind direction/speed, about 6 seconds +/-. Too many variables involved for an exact answer.

Answer 4

How much does it weigh and what was the air temperature and humidity?

Answer 5

45 m/s sin 45 deg =
31.8 m/s

31.8 / 9.81 =
3.24

3.24 * 2 =
6.5 seconds

if u have no idea wat i just calculated, perhaps go seek some help from ur teacher. it's pretty straight forward stuff

Answer 6

until it hits the ground or until someone catches it...

duh!

Answer 7

about 6.49 sec