How to Solve Projectile Motion Problems: Breaking Motion into Components

[gary_shuford]

man i don't understand this stuff i have a test on thursday and i found some practice problems can anyone help me solve thesei need to learn how to do them i don't have the formulasbut pleasee just tell me which formulas to use i don't want answers pleaseee

1. In 1993, Bill tossed a ball 201.24m. Suppose bill threw the spear at 35.0 degree angle with respect to the horizontal. What was the angular velocity of the spear when it left his hand?

2. A softball is through with an angular velocity of 14.5m/s. If you throw the ball at a 35.0 degree angle, calculate the height of the ball at its highest point and how far away the ball is horizontally when it reaches its highest point.

3. A catapult launches a stone into the air at 50.0 degree angle with an angular velocit of 60.0 m/s. How long will the stoe have been in the air when its velocity is -25.0m/s?

4.A kangaroo cleared a fence by jumping wit an angular velocity of 8.42m/s at an angle of 55.2 degrees with respect to the ground. If the jump lasted 1.40 s how tall was the fence assuming the kangaroo barely managed to clear it? What was the kangaroos horizontal distplacement for the trip?

5. You shoot at a basketball goal 3.00 m above your head, making the shot. If the ball is in the air a total of 3.50 seconds and is shot with an angular velocity of 21.0 m/s at what angle did you shoot the ball?

please and thank you

 

[learningphysics]

Does angular velocity here mean initial velocity? I haven't heard the term angular velocity used in these types of problems.

 

[gary_shuford]

possibly I am so conused in this class have you seen problems like this

 

[learningphysics]

possibly I am so conused in this class have you seen problems like this

Yeah... ok for the first problem suppose the initial velocity is v... what is the vertical component of the initial velocity?

 

[gary_shuford]

35 degrees?

 

[learningphysics]

35 degrees?

That's the angle... if v is the initial speed... then the vertical component is vsin(35). The horizontal component is vcos(35). Does this make sense?

 

[gary_shuford]

so like sin is vertical , cos is horizontal don't you take the inverse of it to get it like on the cal u would say sin-1(35)??

 

[learningphysics]

so like sin is vertical , cos is horizontal don't you take the inverse of it to get it like on the cal u would say sin-1(35)??

No... the v is like the hypoteneuse of a right triangle... the horizontal component is the adjacent side of the 35 degrees. The vertical component is the opposite side.

 

[gary_shuford]

so how did u know to use sin in thie problem

 

[learningphysics]

so how did u know to use sin in thie problem

Mainly from experience... the idea is to work on the horizontal part, and the vertical part separately. haven't you solved this type of problem before?

 

[gary_shuford]

im behind i missed a couple days and a couple formulas and occasional nap but I am confused i get the concept but i don't know which and when to use what formulas

 

[learningphysics]

im behind i missed a couple days and a couple formulas and occasional nap but I am confused i get the concept but i don't know which and when to use what formulas

ok. suppose you know that the initial vertical speed is vsin(35). Now, how long does the object take to get to its maximum height? You know the vertical acceleration is -9.8m/s^2.

 

[gary_shuford]

see what formula am i using

 

[gary_shuford]

no help??/?

 

[learningphysics]

no help??/?

sorry I had to leave for a while.

Use vf = vi + at

In the vertical direction, the object gets to the maximum height when vf = 0. vi = vsin(35). So find the time it takes to get to the maximum height.

 

[Kushal]

in this kind of problems, you should be breaking the motion into two. it is called resolving and 2 components are obtained. we most of the time use the vertical and horizontal component.

the initial velocity is the vector. it is at an angle 50(question 3). if you break this 60 ms-1 vector, you will get a 60 cos 50 ms-1 horizontal component and a 60 sin 50 ms-1 vertical component.

a hint is that when the angle is betwwen the actual vector and the horizontal, the horizontal component is cos. and when the angle is between the vertical and the actual vector, the vertical component is now cos. the other one will obviously be sine.

these two components have to be treated separately and they have their own distinct motions.

i'm also learning that in college...that's what i understood.