What were the speed and angle of the diver's projectile motion off a 3-m board?

[Cantworkit]

Homework Statement

A diver leaves a 3-m board on a trajectory that takes her 2.5 m above the board, and the into the water a horizontal distance of 2.8 m from the end of the board. At what speed and angle did she leave the board? Book answer is 7.2 m/s at 77 degrees to the horizontal.

Homework Equations

y= y0 + vy0t - 1/2gt^2

y = x tan theta0 - g/2v0^2 cos^2 theta0*x^2

The Attempt at a Solution

If the origin is at water level, then at the top of the dive vy0 = 0,and 1/2/gt^2 = 5.5 m. Therefore, t = 1.06 s. I still have two unknowns, v0 and theta, but only one equation.

 

[cepheid]

Assuming the t = 1.06 s is right (and I haven't checked too closely), then that means it took her that long to travel a horizontal distance of 2.8 m. So, what was her horizontal velocity?

The time value may actually be wrong, because I think you must use the difference between her final height and her initial height to determine t. (Can you see why it doesn't make sense otherwise? She didn't actually go UP 5.5 m).

Also, if her vertical velocity went from whatever it was initially to zero at the top of the dive in that amount of time, what was her initial vertical velocity? (hint: v_y = v_y0 - gt)

If you know both the initial vertical and initial horizontal velocity, then you know the angle.

 

[Cantworkit]

I do not know the initial velocity. That is what I am trying to solve for, as wll as the angle.

 

[cepheid]

I do not know the initial velocity. That is what I am trying to solve for, as wll as the angle.

I know that. That's why I was trying to explain to you how to find them. Maybe you should read what I said more closely.

 

[Cantworkit]

I am still having trouble. If y0 is at the water, then the total y dispacement is 5.5 m--2.5 m above the board and 3 m below the board. I still can't come up with 7.2 m/s for v0 no matter how I solve for it.

 

[Cantworkit]

Okay, I finally got it. Thanks