Question

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 170 ms. the ball starts from rest. Through what distance does it move before its release? what are the magnitude and direction of the force of the pitcher exerts on the ball?

Answer 1

what's the unit for that time interval?

Answer 2

milisecond i suppose

Answer 3

i got the magnitude of F = 23.88N

Answer 4

explain?

Answer 5

gravitational force acting on the ball is F = mg = 2.21 N........from this we can get m.
then we've been given,
v = 18 m/s, u = 0, t = 170 * 10^(-3) sec.
sub in v = U + at...........this way we can get a.
and F = ma.

Answer 6

hmm

Answer 7

but it's looking for distance right?

Answer 8

i know, i started with the second part pf the question. it seemed easy!

Answer 9

hmm okay..

Answer 10

i got 106 m/s^2

Answer 11

yea, me too.

Answer 12

i got F = 23.88 N

Answer 13

so that's the magnitude?

Answer 14

i think so. and its direction should be horizontal because that's the direction of acceleration.

Answer 15

but positive or negative?

Answer 16

positive.

Answer 17

hmmm so what's the first question

Answer 18

still thinking about that...

Answer 19

do u have any way of confirming the answers?

Answer 20

@mukushla

Answer 21

I'm not too sure abt this:
should the distance traveled before its release be 0?
we know W = F.s
also W = change in KE = 1/2 m (v^2 - u^2)
but the question says "the ball starts from rest". so the change in velocity = 0
thus W = 0.
hence, s = 0.
what do u think?

Answer 22

is the motion look likes this?