Projectile to hit x,y from x,y given max height

In summary, Adrian was having trouble understanding how to solve for the initial velocity that results in a maximum height of 100. After some research, she found that there are several methods that can be used. The first step was to write y = ax^2 + bx, where y(x=0) = 0. From here, she found that b can be found using equation 2. She then set 65 = a*(190)^2 + b*(190), and solved for a. Finally, she found v_x and v_y using the max height and the equation for b.
  • #1
Ehdrian
2
0
Hello!

This is not homework :-)

I can figure this out:
Code:
                                   y = 100 = max height
0,0                                                  200,0
------------------------------------------------------------

By getting the initial y velocity = sqrt(2 * gravity * max height)
and time to peak height = initial_y_velocity / gravity
and the initial x velocity = distance_x / (time_to_peak_height * 2);

However, I need to hit a point with elevation such as:
Code:
                           y = 100 = max height
                                           190,65
                                           -----------------

0,0                                                                         
-------------------------------------------

So, given max height, start point, and end point. What is my initial velocity x, and initial velocity y? No wind, and gravity is constant.

I've really come to a stand still on this! Any help would be great.

Adrian
 
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  • #2
I had a little bit of trouble understanding this but I'm assuming you want the initial velocity that gives a max height of 100 and crosses the point (190,65).

There are probably several approaches but here is what comes to mind:

1) Write y = ax^2 + bx (no need for an extra + c since y(x=0) = 0)

2) Note that y_max = 100 so y(x=-b/2a) = 100. This gives an equation for b in terms of a.

3) set 65 = a*(190)^2 + b*(190), substitute b with the result of step 2. This gives you the value of a, and from part 2, you can find b.

4) Now you know the path exactly, from here, you can v_x by considering the max x distance (-b/a). You already know v_y from the max height.

Remember that there are two solutions!

Does this help?
 
  • #3
Ehdrian said:
Hello!

This is not homework :-)

I can figure this out:
Code:
                                   y = 100 = max height



0,0                                                  200,0
------------------------------------------------------------

By getting the initial y velocity = sqrt(2 * gravity * max height)
and time to peak height = initial_y_velocity / gravity
and the initial x velocity = distance_x / (time_to_peak_height * 2);

However, I need to hit a point with elevation such as:
Code:
                           y = 100 = max height
                                           190,65
                                           -----------------

0,0                                                                         
-------------------------------------------

So, given max height, start point, and end point. What is my initial velocity x, and initial velocity y? No wind, and gravity is constant.

I've really come to a stand still on this! Any help would be great.

Adrian

It is still homework/coursework-like, so it belongs in the HH forums, where I've moved it.

Also, the same rules apply -- you must show us your attempt at a solution, and list the relevant equations. Please show us that work so that we can offer some tutorial advice.
 

Related to Projectile to hit x,y from x,y given max height

1. How do I calculate the initial velocity needed for a projectile to hit a specific point at a given maximum height?

The initial velocity needed for a projectile to hit a specific point at a given maximum height can be calculated using the following formula: v = √(g * (2 * h + 2 * d * tanθ)) where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), h is the maximum height, d is the horizontal distance, and θ is the angle of launch.

2. What is the maximum height that a projectile can reach when launched at a specific angle and velocity?

The maximum height that a projectile can reach can be calculated using the following formula: h = (v * sinθ)² / (2 * g) where h is the maximum height, v is the initial velocity, and θ is the angle of launch.

3. How do I adjust the angle of launch to hit a target at a specific distance?

The angle of launch can be adjusted by using the following formula: θ = arctan((v² ± √(v⁴ - g(g * d² + 2 * h * v²))) / (g * d)) where θ is the angle of launch, v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), d is the horizontal distance, and h is the maximum height.

4. What are the different factors that affect the trajectory of a projectile?

The trajectory of a projectile is affected by several factors, including the initial velocity, angle of launch, air resistance, and the acceleration due to gravity. Other factors such as wind, temperature, and air density can also have an impact on the trajectory.

5. Can I use the same formulas for calculating the trajectory of a projectile in different environments (e.g. on Earth vs on the Moon)?

While the basic formulas for calculating the trajectory of a projectile remain the same, factors such as air resistance and the acceleration due to gravity may differ in different environments. Therefore, adjustments may need to be made to the formulas depending on the specific conditions of the environment.

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