Projectile montion Problem given the Inital Velocity and Angle to horizontal


by keithc
Tags: angle, horizontal, inital, montion, projectile, velocity
keithc
keithc is offline
#1
Dec6-12, 04:25 AM
P: 4
Hi All,
I am struggling to work out the following:
Q: A football is kicked at an angle of 32.50 to the horizontal, with an initial velocity of 42m/s. Assume air resistance is negligible.
A) Find the Balls Velocity at max Height
B) The Balls Acceleratoin Vector at max Height

I have so far worked out the following:
The balls Max height has worked out to be 25.68 m
The Travel time is 8.57 s
and the Distance traveled is 303.5494m
But not sure where to go with this next to figure out the A) and B) parts.
I would very grateful for any help on this.
Thx, K
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grzz
grzz is offline
#2
Dec6-12, 04:42 AM
P: 939
If the method of solution is shown one can give help in a more suitable way.
mrspeedybob
mrspeedybob is offline
#3
Dec6-12, 05:27 AM
P: 689
You don't need any of the 3 things you have figured out in order to get the 2 that you need. In fact, the 3 that you have figured out are quite a bit more complicated then the 2 you are asking about. You are WAAYY over thinking it.

keithc
keithc is offline
#4
Dec6-12, 05:34 AM
P: 4

Projectile montion Problem given the Inital Velocity and Angle to horizontal


Hi All, thanks for looking and inputs. I am new to physics hence I have probably asked a silly question. But if don't ask I will never learn..
Thx again, K
mrspeedybob
mrspeedybob is offline
#5
Dec6-12, 06:00 AM
P: 689
So did you figure it out?
keithc
keithc is offline
#6
Dec6-12, 06:05 AM
P: 4
Not yet !! Still hitting my head of this ball.. hehe.. I am lost as to where to go with my calc's. If I do I will post how I got there hopefully!!
mrspeedybob
mrspeedybob is offline
#7
Dec6-12, 06:17 AM
P: 689
Start by breaking your initial velocity down into vertical and horizontal components. What is factors are going to influence each component while the football is in flight?
keithc
keithc is offline
#8
Dec6-12, 06:48 AM
P: 4
I think I have worked out the velocity at the max height. Not too sure if I am correct or not.
The vertical velocity is 0 so that means that the horizontal velocity worked out by
42. cos 32.5 degs= 35.422 m/s is the same as the velocity at max height or am I totally off with this??
Thx K
mrspeedybob
mrspeedybob is offline
#9
Dec6-12, 11:03 AM
P: 689
Right on. Now the acceleration question is the easiest of all. Since we are neglecting air resistance there is only one other force to consider...


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