Projectile Motion (Baseball Question)

  • Thread starter skateraXIX
  • Start date
  • #1
So I've been having trouble with a problem dealing with projectile motion.

Here's the question:

How much will a ball drop if a pitcher throws the ball horizontally with a speed of 36.5 m/s, from 18m away? (Assuming no spin has been imparted on the ball.)

I'm not sure where to even start at and am really stumped.
 
Last edited:

Answers and Replies

  • #2
madmike159
Gold Member
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You need to use the SUVAT equations (google them). For projectile motion question you can break it into 2 parts horizontal and vertical. The horizontal time needs to be calculated using v=s/t, then you need to work out how much gravity pulls it down in that time. g = 9.81ms^-2.
 
  • #3
Awesome I figured it out. Thanks.

v=d/t
36.5=18/t
Divide 18 by 36.5= 0.493

(v1+v2)/2= (0+4.831)/2= 2.416

Here's how I got the v1 and v2:
v1= 0
v2= g x t
v2= 9.8m/s x 0.493= 4.831

Finally:

d=v x t
d=2.416 x 0.493
d= 1.191 m
 

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