Quick Physics help?

Question

Can somebody break this down for me step by step so I can learn how to do it?

A major league pitcher can throw a baseball in excess of 41.1 m/s. if a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 17.6 m away from the point of release?

Answer 1

1) First, you figure out how long the ball was in the air, while it was being thrown from the pitcher, 17.6 m away to the catcher.

2) Figure out how much the ball dropped in that time, due to gravity.


v = dx/dt
dt = dx/v
dt = 17.6 m/41.1 m/s
dt = 0.428 s

The ball was in the air for 0.428 seconds. Now, we can use this to find out how much it dropped (in terms of elevation):

y = yo + voyt + 0.5at²

y = 0 + 0 + 0.5(-9.80)(0.428)²

y = -0.897 m

*** The ball dropped 0.897 meters by the time the catcher caught it.

Answer 2

OK, here is how it is done. First, some blatant assumptions. Assume there is no air resistance, and the ball has constant velocity. Assume it is thrown horiziontally, and all of it's downward movement is caused by gravity.

So you have a baseball travelling at 41.1 m/s over a distance of 17.6 m Divide to get seconds.

The acceleration of gravity is 9.8 m/s/s.
multiply 9.8 by the number of seconds to get the downward velocity. The average downward velocity would be half that. Multiply the average downward velocity by secords to get the amount that it moved downward.

Answer 3

The forward speed of the ball and it's vertical drop can be considered independently. So, first ask yourself, at that speed, how many seconds is the ball in the air? Secondly, if you merely dropped the ball, how far would it get in that Time under the constant acceleration of gravity (-9.8m/s/s)?

Answer 4

i can solve it if i take a look at my book but you for these kind of problems you should fist learn all the formulas in this topic
then try to understand what you have been given and what you want from all of these.

sorry
goodluck