Solving Projectile Motion Problems: Tips and Tricks for Physics Students

[picklepie159]

Homework Statement

So yesterday, after being confused about two dimensional motion for eons, my physics teacher finally told me that variables could be substituted in other equations. That was when something went CLICK in my mind, and I went home, confident that with this new realization, I could beat any physics question

Easier said than done.

I am still having trouble, even with the simple ones. Here they are-

1.
A man jumps a maximum horizontal distance of 3 meters. How far can he jump on the moon, where downward acceleration is g/6?

#2
A 2 meter tall basketball player, 10 meters away from the hoop 3.05 meters high, launches a basketball at a 45 degree angle. What initial velocity must he use to get it in the hoop without touching the rim?

Homework Equations

Vx= vintial*(cos theta)
X-displacement= Vx*T
V-y final= V-initial (sin theta) + AT
X displacement= V * T

The Attempt at a Solution

For number one-
I split it in half- so now I'm trying to solve
A ball is rolled off the edge of a table and lands 1.5 meters away at 45 degrees. What is Vx,
Vy final, T, and height of the table?

So first, I use
Height= 1/2 g t^2
H= 4.9 *t^2

Since D= Vx * T,
Then T= D/vx
So H= 4.9 * D^2/vx^2
Tan 45= Vy/Vx,
Vx= Vy/tan45
D= 1.5 meters
So H= 4.9 * 2.25/ (vy/tan45)
tan45=1
H= 4.9* 2.25/vy^2

and this is where i stop.

For the second question, I was thinking of perhaps the changing it to

A ball is thrown from ground level, at a 45 degree angle, to a hoop 1.05 degrees.
After that, I have no clue

 

[Karajovic]

hey, here's a good textbook that has a similar problem:
http://www.scribd.com/doc/24324123/McGraw-Hill-Ryerson-High-School-Physics-12-v2

page 68, but in your case, instead, you will have to go backwards and solve for the veloctiy, I believe.

Sorry I couldn't write it all out, I don't have the most of time. Hope it helps.

Take care.

 

[picklepie159]

hey, here's a good textbook that has a similar problem:
http://www.scribd.com/doc/24324123/McGraw-Hill-Ryerson-High-School-Physics-12-v2

page 68, but in your case, instead, you will have to go backwards and solve for the veloctiy, I believe.

Sorry I couldn't write it all out, I don't have the most of time. Hope it helps.

Take care.

just wodering if you are referring to problem one or problem 2?

 

[Karajovic]

just wodering if you are referring to problem one or problem 2?

hey sorry, problem 2

 

[picklepie159]

Ok, so for number two, I got up to this
Y-displacement= Vi*T + (-4.9) t^2

1.05= Vi*T +(-4.9) t^2
10 meters (x-displacement) = Vx * T
T= 10/Vx

1.05= Vy * (10/Vx) + (-4.9)(10/Vx)
tan45= Vy/Vx
Vx=Vy/tan45
Vx=Vy
1.05= 10 +(-4.9) (10/Vx)

I can't think anymore
Can anyone help, please?

 

[The legend]

you'r almost done...

the way is V has 2 components
Vx = V cos45 and Vy=Vsin45
both of which you know to be equal...
you have 2 equations
1.05= Vy*t +(-4.9) t^2 -----------(1)
and
10 meters = Vx * T -----------(2)

so get t in terms of Vx = V cos45 and just substitute in (1) and solve...