Projectile Launched at an Angle: Velocity

[hawkeye1029]

Homework Statement

A bridge rises 321 m above the water. Supposed Jack kicks the rock horizontally off the bridge. The magnitude of the rock's horizontal displacement is 45.0 m. Find the speed at which the rock was kicked.

Homework Equations

Kinematics I think?
v = v_o + at
Δx = v_ot + 0.5at²
v² = v_o² + 2aΔx

The Attempt at a Solution

Well, to start off I tried to divide it into x and y components:

x direction
v_o = ?
v = ?
Δx = 45
a = 0
t = ? t found: 8.09 sec

y direction
v_o = 0
v = ?
Δx = 321 m
a = 9.8 m/s²
t = ? t found: 8.09 sec

So then I tried finding t (which would be the same for both):
321= 0.5(9.8)t²
t = 8.09

But what to do know? What would be the initial velocity of the rock?
Please help, thanks everyone!

 

[SteamKing]

Homework Statement

A bridge rises 321 m above the water. Supposed Jack kicks the rock horizontally off the bridge. The magnitude of the rock's horizontal displacement is 45.0 m. Find the speed at which the rock was kicked.

Homework Equations

Kinematics I think?
v = v_o + at
Δx = v_ot + 0.5at²
v² = v_o² + 2aΔx

The Attempt at a Solution

Well, to start off I tried to divide it into x and y components:

x direction
v_o = ?
v = ?
Δx = 45
a = 0
t = ? t found: 8.09 sec

y direction
v_o = 0
v = ?
Δx = 321 m
a = 9.8 m/s²
t = ? t found: 8.09 sec

So then I tried finding t (which would be the same for both):
321= 0.5(9.8)t²
t = 8.09

But what to do know? What would be the initial velocity of the rock?
Please help, thanks everyone!

You were given how far the rock traveled in the x-direction after it was kicked. You calculated the time it took for the rock to fall from the bridge and hit the ground below.

What else do you need to calculate the speed of the rock when it was kicked?

 

[Mister T]

Δx = v_ot + 0.5at²

You need to apply this equation to both the vertical and the horizontal directions.

In the vertical direction we have

$\Delta y=v_{oy}t+\frac{1}{2}a_yt^2$.

And in the horizontal direction we have

$\Delta x=v_{ox}t+\frac{1}{2}a_xt^2$.

321= 0.5(9.8)t²

Can you show how this follows from the equation for the vertical displacement?

Then you use the value of $t$ you got using this equation in the equation for the horizontal displacement.

 

[hawkeye1029]

OK, and what would be the initial velocity after that?

 

[hawkeye1029]

Would I have to find the square root of the two velocity scalars added together?

 

[hawkeye1029]

Or would the initial velocity be the initial velocity of the x component?

 

[azizlwl]

You have initial vertical velocity, its acceleration and distance traveled. The rock will travel forward until it is stopped by the water.

 

[hawkeye1029]

Thanks for all the replies :).
So how would I use all this information to get the overall initial velocity? Is there some kind of formula or would I have to calculate it in some other way?

 

[SteamKing]

Or would the initial velocity be the initial velocity of the x component?

Well, the rock is launched horizontally off the bridge, according to the problem statement.

You've done all the heavy lifting in this problem. All that remains is to calculate the horizontal velocity.

 

[hawkeye1029]

Ohh so the horizontal initial velocity is the answer to the problem?

 

[SteamKing]

Ohh so the horizontal initial velocity is the answer to the problem?

Well, sometimes the most important part of solving a problem isn't making all the fancy calculations. It's reading and understanding what the problem wants.

When in doubt, re-read the problem statement.

 

[hawkeye1029]

OK :)
TY all !