Mechanics question. Person jumping off trampoline.

[Educated]

Homework Statement

Alex jumps off the edge of a trampoline at an angle of 40 degrees to the horizontal. The trampoline is 0.85 meters above the ground and Alex is in the air for 0.64 seconds.
(I did a rough sketch of it)

Calculate how far away from the edge of the trampoline Alex lands.
Neglect air resistance.

Homework Equations

The simple kinematic equations:
$$d = v_i t + \frac{1}{2}at^2$$

$$v_f = v_i + at$$

$$v_f \, ^2 = v_i \, ^2 + 2ad$$

$$d = \frac{(v_i + v_f)t}{2}$$

The Attempt at a Solution

I don't even know where to begin. The part that really confuses me is that the upwards part of the travel is shorter than the downwards part of the travel. I cannot solve the initial velocity, final velocity nor the distance.

Can someone just give me a hint in the right direction. I don't want the answer, leave me to solve the answer myself.

 

[Nynex]

Let the y traveled equal to -0.85...express x in terms of velocity...?

 

[Educated]

I'm sorry, what do you mean?
This question is really confusing me.

What is y and what is x? What will expressing things in terms of x and y achieve?

Mechanics really isn't my strong point.

 

[Nynex]

Have you guys studied projectile motion in class? The equations in your original post are based on a single dimension of displacement. In this problem Alex moves in two dimensions, those equations won't work!

Essentially, y would be the displacement in the y direction (up and down) and x would be the displacement in the x direction (left and right).

Alex displaces to the right from his original position. He also displaces down 0.85 m, since he lands on the ground below his starting position.

Using these two equations we should be able to solve for Alex's displacement in the x direction.
(sorry unfamiliar with latex)

x = (v*cos O)*t
y = (v*sin O)*t - g/2*t^2

 

[rwisz]

Hi there,

I would approach this problem by first identifying the key information and variables that we have been given. We know that Alex jumps off the edge of a trampoline at an angle of 40 degrees to the horizontal, and that the trampoline is 0.85 meters above the ground. We also know that Alex is in the air for 0.64 seconds and that we can neglect air resistance.

From these given values, we can use the kinematic equations to determine the distance that Alex lands from the edge of the trampoline. The key equation to use is d = v_i t + \frac{1}{2}at^2, where d is the distance, v_i is the initial velocity, t is the time, and a is the acceleration.

To begin, we can calculate the initial velocity using the equation v_f = v_i + at. Since we know that Alex is in the air for 0.64 seconds, we can plug this value in for t and solve for v_i. Remember that the acceleration due to gravity is always -9.8 m/s^2.

Once we have the initial velocity, we can use it in the equation d = v_i t + \frac{1}{2}at^2 to calculate the distance that Alex travels while in the air. However, we need to be careful with the angle of 40 degrees - this means that the initial velocity has both a horizontal and vertical component. We can use trigonometry to determine the horizontal and vertical components of the initial velocity, and then use these values in the equation to find the distance traveled.

To find the distance that Alex lands from the edge of the trampoline, we can use the equation d = \frac{(v_i + v_f)t}{2}. Again, we will need to use the horizontal and vertical components of the initial velocity, along with the time and acceleration, to solve for the distance.

I hope this helps guide you in the right direction. Remember to carefully consider the given information and choose the appropriate equations to use. Good luck!