Projectile Motion Problem: Bullet and Baseball Examples

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Homework Help Overview

The discussion revolves around two projectile motion problems involving a bullet and a baseball. The first problem concerns a bullet traveling horizontally and determining how far it falls before hitting a target. The second problem involves a baseball hit at an angle, with questions about its time in the air and horizontal distance traveled.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the time of flight for the bullet and the vertical distance it falls. They also discuss the horizontal and vertical components of the baseball's motion, including its time in the air and horizontal distance.

Discussion Status

Some participants have provided hints and guidance regarding the equations of motion and the relationships between time, distance, and acceleration. There is an ongoing exploration of the assumptions and calculations involved in both problems, with no explicit consensus reached.

Contextual Notes

Participants note potential typos in the values for gravitational acceleration and clarify the correct horizontal velocity for the bullet. There is also mention of the need to visualize the projectile's path for better understanding.

willingtolearn
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#1
A bullet traveling 800 m/s horizontally hits a target 180 m away. How far does the bullet fall before it hits the target ?
---------------------------------------
My work:
X
d = 180 m
Vx = 180 m/s
t = .225 s

Y
a = -9.8 m/s2
t = .225 s (time in x direction = time in y direction)
I'm stuck here ! Help please



#2
A baseball was hit at 45 m/s at an angle 45' above the horiztontal .
a. HOw long did it remain in the air ?
b. How far did it travel horizontally ?
-------------------------------------

My work:
Vx = 31.8 m/s = Vy
ay = -9.8 m /s2
I only got to that point ! Help please
 
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Here's a hint - a body being accelerated at a ms^-2 travels distance s = 0.5at^2 m in time t.

[tex]s = \frac{1}{2}at^2[/tex]
 
Mentz114 said:
Here's a hint - a body being accelerated at a ms^-2 travels distance s = 0.5at^2 m in time t.

[tex]s = \frac{1}{2}at^2[/tex]

for problem #1 or #2?
 
In problem #1, you know how long the bullet was in the air ... which is enough to solve the problem. Do you know that a falling body accelerates at g = 32 m/sec/sec ?
 
problem 1: vx=800 m/s not 180m/s
time is correct.

think about this:
a=-9.8 m/s^2
t=.225
vo=0
x=volt+.5at^2, as mentz114 stated: vo=0 so x=.5at^2
you have a and t, now find x

problem 2:
vx=vy=31.8m/s
ay=-9.8n/s^2
ball is hit? can you visualize the arch that represents the ball in the air?
ball has to travel from hit point to maximum height
vf=vo+at, as maximum height, what is vf? once you have that, solve for time
this time is the time from hit to max height, what do you need to do to find time from hit to max to horizontal?
when you find the time from hit to max and back to horizontal, d=vt, solve
 
Wow ... it look really easy ! Thanks a lot silvashadow, Mentz114
I got them all !
 
Mentz114 said:
In problem #1, you know how long the bullet was in the air ... which is enough to solve the problem. Do you know that a falling body accelerates at g = 32 m/sec/sec ?

It's probably a typo, but g is 9.8m/s/s not 32 :)

These make it look so easy! Gaaah
 
willingtolearn said:
#1
A bullet traveling 800 m/s horizontally hits a target 180 m away. How far does the bullet fall before it hits the target ?
---------------------------------------
My work:
X
d = 180 m
Vx = 180 m/s
t = .225 s

Y
a = -9.8 m/s2
t = .225 s (time in x direction = time in y direction)
I'm stuck here ! Help please
You have it correct now just find the vertical distance... y=1/2gt*2.. with the time you found.


willingtolearn said:
#2
A baseball was hit at 45 m/s at an angle 45' above the horiztontal .
a. HOw long did it remain in the air ?
b. How far did it travel horizontally ?
-------------------------------------

My work:
Vx = 31.8 m/s = Vy
ay = -9.8 m /s2
I only got to that point ! Help please

So far you are correct... No solve how long it would take the ball to travel straight up at that velocity. t= vf-vi/g... that is to the maximum height... so you must double it for the time up and down.

part b you now have the time it was in the air so now use... d=Vx X T... Make sure you use the total time you just solved for.
 

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