2D projectile motion (on calculator?)

[Jst2getItdone]

Homework Statement

A person is standing on a cliff that is 10m tall, the boy is 2m back from the ledge of the cliff.
He shoots an arrow at a 45 degree angle. Initial velocity is 5 m/s.

I need to find max height when the arrow will reach the height. How far the arrow went in the X direction, and How far should I build a wall that's .5m high, so that the arrow just misses the wall.

**We can use our calculators.

Homework Equations

To set this up I used y1= -9.8 y2=-9.8t y3=10+3.5t-4.9x^2 y5=3.5 y6=-2+3.5t
I found 3.5 from the Sin and Cos of 45.

The Attempt at a Solution

I found the max height and the time that the arrow reached it from the parabola on my calculator 10.6m at .36sec
but I can't seem to find the other ones. I just need to know which line to look at.

Thanks so much!

 

[anathema]

Thank you for your question. The first thing we need to do is to define the variables in this problem. Let's use h for the height, x for the horizontal distance, and t for time. We can also define g as the acceleration due to gravity, which is equal to -9.8 m/s^2.

Now, let's look at the motion of the arrow in the vertical direction. We know that the initial vertical velocity is 5sin(45) = 3.5 m/s. Using the equation h = h0 + v0t + 1/2at^2, we can find the height at any time t. Since the arrow starts at a height of 10 m, we can set h0 = 10. This gives us the equation h = 10 + 3.5t - 4.9t^2.

To find the maximum height, we need to find the vertex of this parabola. We can either use the formula for the vertex (x = -b/2a, y = c - b^2/4a) or we can use calculus to find the maximum value of h. Either way, we get the same result: the maximum height is 10.6 m, reached at a time of 0.36 seconds.

Next, let's look at the horizontal motion of the arrow. We know that the initial horizontal velocity is 5cos(45) = 3.5 m/s. Using the equation x = x0 + v0t, we can find the horizontal distance at any time t. Since the arrow starts at a horizontal distance of 2 m, we can set x0 = 2. This gives us the equation x = 2 + 3.5t.

To find the distance the arrow went in the x direction, we need to find the value of x at the time when the maximum height is reached. From the previous calculation, we know that this happens at t = 0.36 seconds. Plugging this into the equation, we get x = 2 + 3.5(0.36) = 3.7 m.

Finally, let's look at the horizontal distance the arrow needs to travel to just miss a wall that is 0.5 m high. We can set up a similar equation as before, but this time we need to find the time when the arrow reaches

 

[kerimek]

I would approach this problem by first setting up the equations of motion for a projectile in a two-dimensional plane. From the given information, we know that the initial velocity of the arrow is 5 m/s and it is launched at a 45 degree angle. Using this information, we can find the initial velocities in the x and y directions, which would be 5*cos(45) and 5*sin(45), respectively.

Next, we can use the equations of motion to find the maximum height reached by the arrow. This can be done by setting the vertical position equation (y) equal to zero and solving for the time at which the arrow reaches its maximum height. From there, we can plug that time back into the horizontal position equation (x) to find the horizontal distance traveled by the arrow.

To find the distance at which the arrow just misses a 0.5m wall, we can set the vertical position equation equal to 0.5m and solve for the horizontal distance. This will give us the minimum distance the wall should be from the cliff.

Using these equations and principles, we can solve the problem without relying on a calculator. However, if you choose to use a calculator, the method described in the attempt at a solution seems to be on the right track. It may be helpful to plot all the equations on the same graph and look for the intersection points to find the relevant information.