if i throw a ball up at 45 m/s. how long is the ball in the air?

Answer 1

Final velocity "V" equals initial velocity "u" plus acceleration "a" times time "t".

v = u + at

Here v = 0, so 45 = 9.8t or t = 45/9.8 = 4.6 seconds.
----------- ---------- ------------ ----------- -------------- ----------------
EDIT: doug_donaghue (answerer below) is right, the ball is in the air for 9.2 seconds, I totally forgot the falling time, which should be exactly the same if air resistance is ignored.

Answer 2

Hello Betsy, time of flight in case of oblique projection is given by the expression 2 u sin @ / g. Here @ is the angle of projection. u- the velocity with which it is projected and g-acceleration due to gravity Since the ball is thrown vertically upward @ = 90 deg So the required time = 2 * 45 / 9.8 = 9.18 s

Answer 3

I get 9.18 seconds.
You must have one helluva good arm ☺
That's right at 100 mph.

Doug
EDIT: How long does it have to fall at 9.8 m/s² to get to 45 m/s? Bout 4.56 s. Good. Because that's -exactly- how long it took to get up there as well. So it was in the air a -total- time of 9.18 s.

Answer 4

it would take a total of 9.2 seconds

4.6 seconds to up and then 4.6 seconds to come down

Answer 5

45/9.81 is what it comes down to.

4.59 seconds.

that is assuming you throw it straight upwards, and not at any angle.

Answer 6

Where are you standing? on a cliff? on a tree ? the ball is falling into your hand? or on the floor? Are you in space? All these information are required, please.

Answer 7

45/9.8 it is about 4.6 seconds.

Answer 8

45m/s divided by 9.8 m/s^2
that should give u the time