Solving Projectiles: Kinematics Equations & Variables for Grade 12 Physics 12 BC

[link2110]

Hi I am a grade 12 student in Physics 12 in BC, and I was wondering for projectiles, what equations you use, and what variables you use to solve. For example, in our notes our teacher gave us and example and completed it for us, but I don’t get why he used the equations he did. I do know that you use kinematics equations, and you have to change the initial velocity of the object into its 2 components V_y and V_x using component method, but other than that I don’t get why you use an equation and not another one.

Here is the problem, and any help at all would be appreciated.


A golf ball is hit diagonally (37^o) off the top of a 15m cliff at 40m/s.

Find:
a) Time to max height
b) Max height achieved
c) Entire time in air
d) Range
e) Velocity at 1.0 seconds
f) Velocity at impact


Here is the picture that I drew to start.

http://img23.imageshack.us/img23/556/physicsquestion.th.png

Then I found the V_x and the V_y.

http://img3.imageshack.us/img3/1953/physicsquestion1.th.png

Now I'm stuck to find the first thing...because I don't know how to get it.
Help please?




The answers to the question are:
a) 2.46 sec
b) 44.6 m
c) 5.47 sec
d) 175 m
e) 35.0 m/s @ 24.1^o above horizontal
f) (do not have b/c notes are incomplete)

 

[slider142]

The key to these problems is in the assumption that each independent dimension of motion is unaffected by the other. That is to say, if, parallel to the x-axis, the ball has no horizontal acceleration, then you can use kinematics equations assuming that the ball has no acceleration and get the correct horizontal component of the ball's velocity irrespective of what is going on in the y-direction. The same holds for the y-component analysis.

So if we look at question a) [time to max height], realize that the height of the ball has nothing to do with the horizontal velocity of the ball, and thus we need only know the y-component of the velocity and whether the ball is accelerating in that component. Use trigonometry to get v_y and note that the acceleration in y is -g, where g is the scalar acceleration due to gravity at sea level. In the y-direction, the projectile is simply going up and down, just as if you were to throw it straight up with velocity v_y without any horizontal component. You should already know the proper equation for this setup from your study of one-dimensional kinematics.

 

[link2110]

Use trigonometry to get v_y and note that the acceleration in y is -g, where g is the scalar acceleration due to gravity at sea level. In the y-direction,

so I am assuming that's -9.80 m/s - accelleration due to gravity.

then i'd use an equation just like the ones in kinematics where the ball is thrown straight up (V2= V1 + at), and assume that at V2, the max height the velocity of the ball is 0m/s. the V1 is the inital accell of the ball, which is the Vy. Am i right?

 

[slider142]

so I am assuming that's -9.80 m/s - accelleration due to gravity.

then i'd use an equation just like the ones in kinematics where the ball is thrown straight up (V2= V1 + at), and assume that at V2, the max height the velocity of the ball is 0m/s. the V1 is the inital accell of the ball, which is the Vy. Am i right?

Sounds good! You will also need an equation involving distance, as you need height.

 

[link2110]

how do you find the max height?

 

[slider142]

Use an equation relating distance to acceleration and time. Find the time it takes for the object to decelerate to 0 using the equation you gave before. When it reaches 0 velocity it has attained its maximum height and will start falling (negative velocity).