Deriving Projectile Motion Equations

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SUMMARY

The discussion focuses on deriving equations for projectile motion involving a projectile launched from ground level at an angle θ with an initial speed V0. The solutions provided for each part are as follows: Part A calculates the time to maximum height as t = (V0sinθ)/g; Part B finds the total time until the projectile hits the ground as t = (2V0sinθ)/g; Part C determines the maximum height as y = (V0sinθ)²/(2g); and Part D computes the horizontal distance R traveled as R = (V0²sin2θ)/g. The calculations are confirmed to be correct, particularly for Part C, despite initial doubts.

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  • Knowledge of gravitational acceleration (g)
  • Ability to manipulate algebraic equations
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Homework Statement



A projectile is fired from ground level at time t=0, at an angle of θ with respect to the horizontal. It has initial speed V0.

Part A

Find the time it takes the projectile to reach its maximum height. Express in terms of V0, θ, and g.

Part B

Find the time at which the projectile hits the ground after having traveled through a horizontal distance, R. Express in terms of V0, θ, and g.

Part C

Find the maximum height attained by the projectile. Express in terms of V0, θ, and g.

Part D

Find the total distance R traveled in the x direction. Express in terms of V0, θ, and g.


Homework Equations



V=V0+at

y=y0+Vy0t+0.5ayt2

x=Vx0t

The Attempt at a Solution



For Part A, I did:

Vy=Vy0 + ayt

0=V0sinθ-gt

gt=V0sinθ

t=(V0sinθ)/g

For Part B I did:

y=y0+Vy0+0.5ayt2

0=0 + (V0sinθ)t- 0.5gt2

0.5gt2=(V0sinθ)t

gt2= (2V0sinθ)t

t= (2V0sinθ)/g

For Part C I did:

Vy2=Vy02+2ay

0=(V0sinθ)2-2gy

y= (V0sinθ)2/2g

For Part D:

x = x0 + Vx0t

x= V0cosθ × (2V0sinθ)/g

x= (V02sin2θ)/g

Did I do this correctly?? Or did I just mess up everything.. The one I'm really confused about is Part C.. I don't know if I could use the equation that I used..
 
Physics news on Phys.org
All good.
 
Really!? So I am right.. I was told by someone that part C was wrong.. o_o
 
Checked it again... part C still looks right to me.
 

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