Finding initial velocity of a free-falling object that launched off a cliff

[kmb11132]

Homework Statement

A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's displacement 0.77 s after leaving the dock has a magnitude of 7.4 m. What is the car's speed at the instant it drives off the edge of the dock?

@t = .77s displacement is 7.4m
a = -9.8m/s^2

Homework Equations

1. v_f^2 = v_i^2 + 2ad

or

2. d = v_i * t + (a * t^2)/2

The Attempt at a Solution

v_f = 7.4/.77 = 9.61m/s

1. 9.61^2 = v_i^2 + 2 * -9.8 * 7.4

... v_i = 13.38m/s, incorrect (aware of incorrectly mixing x and y components)

2. 7.4 = v_i * .77 - (9.8 * .77^2)/2

... v_i = 13.38m/s, incorrect (aware of incorrectly mixing x and y components)

3. Making a right triangle by finding v_f of an initially stationary object falling for .77s:

at + v_i = v_f

-9.8 * .77 = v_f

v_f = -7.55m/s

9.61^2 = 7.55^2 + b^2

b = 5.95

so v_i = 5.95m/s, incorrect

 

[Ignea_unda]

Homework Statement

A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's displacement 0.77 s after leaving the dock has a magnitude of 7.4 m. What is the car's speed at the instant it drives off the edge of the dock?

@t = .77s displacement is 7.4m
a = -9.8m/s^2

Homework Equations

1. v_f^2 = v_i^2 + 2ad

or

2. d = v_i * t + (a * t^2)/2

The Attempt at a Solution

v_f = 7.4/.77 = 9.61m/s

Umm, I think this might be where you got off a little. Displacement is not the same as velocity. Displacement is a distance.

Try using another of the equations for uniformly accelerated motion to work on this one.

 

[kmb11132]

Umm, I think this might be where you got off a little. Displacement is not the same as velocity. Displacement is a distance.

Try using another of the equations for uniformly accelerated motion to work on this one.

So what I found would be v_av not v_f correct? I'm kind of stuck right now because I don't think I can use equations that incorporate v_f, and those that incorporate v_i, a, d, and t all produce 13.38 as the initial velocity. I don't think I can separate this problem into vertical and horizontal components because I don't have any sort of triangle, but I get the impression that this is somehow what must be done. This is because the acceleration in directly downward, and v_i I'm looking for is directly eastward, and the distance I'm given falls in between these two. Any further hints would be greatly appreciated.

 

[gneill]

A diagram might help!

 

[zgozvrm]

Since you didn't specify the direction of displacement (7.4 m horizontally, vertically, or diagonally from the dock), I would assume diagonal displacement.

 

[kmb11132]

Assuming the car fell at a 45 degree angle the magnitude of its x displacement would be 5.23 meters in .77 seconds. Without an x acceleration/assuming constant speed this would work out to a v_ix of 6.79m/s. This is also incorrect. There is still some key to solving this problem that's missing here and I'm seriously at a loss for what it is. That's a really nice diagram, by the way, gneill.

 

[gneill]

The car won't fall at a 45° angle, it will follow the parabolic path of a projectile. The horizontal velocity will be constant, but the vertical velocity accelerates downwards due to gravity. You need to use the kinematic equations for projectile motion.

 

[kmb11132]

That's what I thought, but I figured in the absence of all other ideas I'd give a 45-45-90 triangle a shot. I tried all of the kinematic equations for projectile motion in every way that I thought made sense and they all incorrectly lead me to a v_ix of 13.38m/s. I know that I'm using the equations incorrectly because I'm mixing vertical and horizontal components. The problem is without an angle I can't figure out how to split the given information into vertical and horizontal components.

 

[gneill]

Given some (as yet unknown) initial horizontal velocity Vo, write the expressions for the X and Y positions of the car with respect to time. For simplicity, take the downward direction to be positive, then the acceleration will be positive, too.

 

[kmb11132]

Thank you so much! I didn't think there was enough information to do that but, well, there definitely was.

For the sake of future Googlers coming to this page what I did was...

Take this equation: d = v_i*t + .5*a*t^2

and solved for the y displacement:

d_y = 0 * .77 + .5 * -9.8 * .77^2

d_y = -2.9m

This gives me a displacement right triangle. Using the Pythagorean theorem I determined that:

d_x = 6.8m

Knowing that this distance was traveled over .77s, and that horizontal velocity is equal throughout free-fall:

v_i = (6.8m / .77s) = 8.83m/s