Question

A ball is thrown straight up in the air with a speed of 30 m/s. How much time does it take for the ball to return to the thrower? The acceleration due to gravity has an exact value of 9.8 m/s2, but you may use 10 m/s2. Hint: what would the final velocity be? Make sure to include the positive or negative in that final velocity

Answer 1

final velocity will be 30 down or -30m/s

to get time up:
(1/2) m v^2 at bottom = m g h at top
so
30^2 = 2 g h
h = 900/(2*10) = 90/2 = 45 meters
average speed up = 30/2 = 15 m/s
so time up = 45/15 = 3 seconds
time down is also 3
so
6 seconds in the air

Answer 2

U=30m/s²
V= 0


V = u -gt

0 = 30 - 10t

t=3s

Answer 3

That was the time up, but they want total time in the air.

Answer 4

We have not learned that second part yet. All we were told was to use the equation Vf=Vo+at and I plugged in
-30=30+10(t) but all I get is -6 not 6

Answer 5

Or would Vf^2+Vo^^2+2a(Xf-Xo) work

Answer 6

but a is down, negative
-30 = 30 - 10 t
10 t = 60
t = 6 seconds

(that is the quickest way to do it. I was trying to lead you through everything that is going on.)

Answer 7

Yes
Or would Vf^2+Vo^^2+2a(Xf-Xo) work
Is really what I used to get the height
v^2 = 2 g h

Answer 8

Thank you!!

Answer 9

You are welcome.

Answer 10

V = Vo + g*Tr = 0.
30 - 9.8Tr = 0,
Tr = 3.06 s. = Rise time.

Tf = Tr = 3.06 s.

Tr+Tf = 3.06 + 3.06 = 6.12 s. = Time in air.