Deriving Projectile Motion Equations

In summary, the conversation discusses the calculation of various parameters for a projectile fired at an angle with initial speed V0. Part A finds the time it takes for the projectile to reach its maximum height, expressed in terms of V0, θ, and g. Part B finds the time at which the projectile hits the ground after traveling a horizontal distance R, also expressed in terms of V0, θ, and g. Part C finds the maximum height attained by the projectile, expressed in terms of V0, θ, and g. Finally, Part D calculates the total distance R traveled in the x direction, expressed in terms of V0, θ, and g.
  • #1
rakeru
75
0

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..
 
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  • #3
Really!? So I am right.. I was told by someone that part C was wrong.. o_o
 
  • #4
Checked it again... part C still looks right to me.
 
  • #5




Your solutions for Parts A, B, and D are correct. However, for Part C, you have used the wrong equation. The equation you used, Vy2=Vy02+2ay, is for finding the final velocity in the y-direction, not the maximum height. To find the maximum height, you can use the equation y=y0+Vy0t+0.5ayt2, where y0 is the initial height (in this case, y0=0) and Vy0 is the initial vertical velocity (in this case, Vy0=V0sinθ). Setting this equation equal to zero and solving for t will give you the time at which the projectile reaches its maximum height. Plugging this back into the equation for y will give you the maximum height in terms of V0, θ, and g.
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It is a combination of horizontal and vertical motion, resulting in a curved path.

2. How do you derive the equations for projectile motion?

The equations for projectile motion can be derived using the principles of kinematics, which is the study of motion without considering the underlying forces. Specifically, the equations are derived by analyzing the horizontal and vertical components of the object's motion separately and then combining them to form a single equation.

3. What are the key assumptions made when deriving projectile motion equations?

The key assumptions made when deriving projectile motion equations are that the object is moving in a uniform gravitational field, there is no air resistance, and the object is small enough that the curvature of the Earth can be ignored.

4. What are the four equations for projectile motion?

The four equations for projectile motion are the equations for displacement, velocity, acceleration, and time. They are as follows:
- Displacement: x = x0 + v0t + 1/2at2
- Velocity: v = v0 + at
- Acceleration: a = -g (constant vertical acceleration due to gravity)
- Time: t = (v - v0)/a

5. How is the range of a projectile calculated?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation R = v2sin(2θ)/g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

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