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Physics Word Problem; Projectile Motion

  • Thread starter georgiaa
  • Start date
  • #1
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Homework Statement



Suppose a catcher is crouched down behind the plate wen he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 ft from the catcher to second base?


Homework Equations



v= d/t
The big 5 equations
sine law
cosine law
pythagorus

The Attempt at a Solution



X = 120 feet
V = velocity at which ball was thrown
Θ = 30 degrees
g = acceleration due to gravity = 32.2 ft/sec^2 (constant)

120 = V^2(sin 2*30)/32.2

V = sqrt (120*32.2/sin 60)

V = 66.80 ft/sec.

T = 120/66.80

T = 1.80 sec.

I'm not sure if this is right...is there an easier solution using the formulas provided above?
 
Last edited:

Answers and Replies

  • #2
Kurdt
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  • #3
LowlyPion
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Homework Statement



Suppose a catcher is crouched down behind the plate wen he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 ft from the catcher to second base?


Homework Equations



v= d/t
The big 5 equations
sine law
cosine law
pythagorus

The Attempt at a Solution



X = 120 feet
V = velocity at which ball was thrown
Θ = 30 degrees
g = acceleration due to gravity = 32.2 ft/sec^2 (constant)

120 = V^2(sin 2*30)/32.2

V = sqrt (120*32.2/sin 60)

V = 66.80 ft/sec.

T = 120/66.80

T = 1.80 sec.

I'm not sure if this is right...is there an easier solution using the formulas provided above?
Looks right (66.7696 f/s) for velocity, But I think you haven't used the horizontal component of velocity. Looks like you used Vo to figure the time.
 
Last edited:
  • #4
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Wow, this is a fun little question. Here's an idea (double check this yourself - don't trust it!):

In the horizontal direction:
d = vt + .5at^2
120 = vt + 0
So horizontal velocity is:
v = 120 / t
So vertical velocity is:
v = (tan30) x horizontal velocity
v = (tan30) x 120 / t

Now in the vertical direction:
d = vt + .5at^2
0 = (vertical velocity x t) + .5(-32.2)t^2
t = ?

Edit - removed complete solution as per forum rules.
 
Last edited:

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