Honors physics basic projectile motion question

[Kalix]

Homework Statement

Question: Inbar and Luc are captured by angry natives on a desert island, after their plane crashes. They need to jump over a pit filled with snakes. They jump up at an angle of 23 degrees and need to land 8.7 meters away in x. What does their take off speed have to be?

The variables given are in the problem. My biggest problem is I don't understand how to combine the x and y equations in projectile motion. Here are the variables I think I have but please correct me if I am wrong.
a=-9.81m/s^2
x=8.7m
Vfy=0m/s
Vi=?
t=?
Is there anything else I need to find or any other variables I should already know from the problem?

Homework Equations

Here are some equations which I have tried using.
Vfy=Viy+ayt
X=Vi,avg t

The Attempt at a Solution

So here is what I have tried. It's not much but it's my small attempt.
First I set up a right triangle trig problem. Then I used tangent to find that viy=3.7. Next I used Vfy=Viy+ayt to try to find t. From here I get kinda stuck. I don't even know if the first part was correct but at least I tried.

 

[Sunny1261]

So this problem requires a little bit of abstract thought.

We know that the x velocity, Vx has to be just enough to clear the snakes, so 8.7/t, where t is your hang time.

By the trig of the problem, we can also say that Vx = V cos(23).

We know that Vy has to be just enough so that produces the desired hang time, t, which equals 8.7/Vx = 8.7/(V(cos(23)).

So we have:

t = 8.7/(Vcos(23))

So need to figure out the y component.

We can use the equation Vf = Vo + at and apply it to the Y kinematics.

So -Vy = Vy - 9.81 * t (We say -Vy is Vf because when he lands, his Y velocity will be opposite but equal his up Y velocity).

From here, we can say Vy = Vsin(23).

We now have our two equations.

t = 8.7/(Vcos(23))
-Vsin(23) = Vsin(23) - 9.81 * t.

Two equations, two unknowns. Happy Hunting.

 

[Kalix]

I am a bit confused about the V in Vsin(23) and Vcos(23). Is the V in x or Y? Is it Vi or something like that?

 

[Sunny1261]

V is the final velocity.

If you draw your triangle, you have V going off at an angle, and Vx horizontal and Vy vertical. So you can say that Vx = V * cos(23) while Vy = V * sin(23).

 

[rwisz]

Hello,

Thank you for sharing your attempt at the solution. It seems like you are on the right track with setting up a right triangle and using trigonometry to find the initial vertical velocity (viy). However, there are a few other variables that you can use to solve this problem.

First, you can use the fact that the initial horizontal velocity (vix) is equal to the final horizontal velocity (vfx) since there is no acceleration in the horizontal direction. This means that vix = vfx = vi*cos(23), where vi is the initial velocity and 23 is the angle at which they jump.

Next, you can use the equation x = vix*t to find the time (t) it takes for them to reach the other side of the pit. You know the value of x from the problem (8.7m) and you just found the value of vix. Plug these values into the equation and solve for t.

Now that you know the time it takes for them to reach the other side, you can use the equation y = viy*t + 1/2*ay*t^2 to find the initial vertical velocity (viy). You know the value of y (which is the height of the pit, but you can assume it to be 0 for simplicity), the time (t) that you just found, and the acceleration due to gravity (ay). Plug these values into the equation and solve for viy.

Finally, you can use the Pythagorean theorem to find the magnitude of the initial velocity (vi). The Pythagorean theorem states that the sum of the squares of the lengths of the sides of a right triangle is equal to the square of the length of the hypotenuse. In this case, the sides are viy and vix, and the hypotenuse is vi. So, you can write the equation vi^2 = viy^2 + vix^2 and solve for vi.

I hope this helps! Let me know if you have any other questions.