How Can Projectile Motion Equations Be Derived from Basic Kinematics?

[madeeeeee]

I want to derive some physics equations from other equations.

a) How can I derive this formula:
R=Vo^2(sin(sα) / -a
from these two equations:
(1) x=(Vo cosα)t
(2) y=(Vo sinα)t + 1/2at^2

Things I know
- I think the way to do this is to isolate the variable t in equation (1) and substitute the t in equation (2).
- Is the identity 2sin(theta) cos(theta)= sin(2theta) useful?
-When the projectile returns to its original level y=0
Please help me understand how to do this.

b) From equation
y=yo + Vyo t + 1/2at^2

how can I show that:
h= -1/2a(tf)^2

If I measure h and tf and recall that
a=-g sin (theta)
than this equation can be used to derive an equation for g in terms of tf, h and theta.

Please help!

 

[LawrenceC]

It would help if you defined your variables. What is sa? What is R? What is tf?

 

[madeeeeee]

sorry its not sa its (2(alpha))
R is the range
tf is final time

so I did the part b and I got g=2h/(sin(theta)(tf)^2)...is this right? I still need help for the a)

 

[LawrenceC]

Hint: Part a

Determine how long the projectile is in the air by determining how long it takes to reach its apex, then double it because it has to come down. You can get an equation for this time by using a little calculus on equation 2.

"Is the identity 2sin(theta) cos(theta)= sin(2theta) useful? "

Yes

 

[paranormal]

a) To derive the projectile equation R=Vo^2(sin(sα) / -a, we can start by solving for t in equation (1) and substituting it into equation (2). This will give us an equation for y in terms of x and α:
y = xtanα - (gx^2)/(2Vo^2cos^2α)

Next, we can use the identity 2sinθcosθ = sin(2θ) to rewrite the equation as:
y = xtanα - (g/2Vo^2)sin(2α)x^2

Now, we can see that the maximum range of the projectile occurs when y = 0. So, we can set the equation equal to 0 and solve for x:
0 = xtanα - (g/2Vo^2)sin(2α)x^2
x = 0 or x = (2Vo^2/g)sin(2α)

Substituting this value of x back into the original equation, we get:
R = (2Vo^2/g)sin(2α)tanα - (g/2Vo^2)sin(2α)(2Vo^2/g)^2
R = (Vo^2/g)sin(2α)(2tanα - sin(2α))

Finally, we can use the identity sin(2α) = 2sinαcosα to simplify the equation:
R = (Vo^2/g)sinα(cosα - sinα)

b) To show that h = -1/2a(tf)^2, we can start by substituting the given values of a and t into the equation:
h = yo + Vyo(tf) + (1/2)(-g sinθ)(tf)^2

Next, we can use the fact that yo = 0 and Vyo = 0 at the highest point of the projectile's trajectory. This will simplify the equation to:
h = (1/2)(-g sinθ)(tf)^2

Finally, we can use the given values of h and tf to solve for g:
g = -(2h)/(sinθ(tf)^2)