Finding initial velocity of an object reaching 10 m high

[Blobikins]

Homework Statement

I need to find two ways to calculate initial velocities (2 versions) of an object just clearing over a 9.99 m wall.

Homework Equations

kinematics equations

The Attempt at a Solution

Way 1 I was trying is to say dx = 40m , dy = 10m, ay= -9.81m/s^2.

The thing is, I can't get started even for y because I'm missing too many variables, namely viy, vfy, and t.

The second way is to say the angle is 45 degrees, so viy is (something)sin45 and vx is (something)cos45.

But again, I'm missing too many variables to calculate it.

Am I missing something?

 

[HallsofIvy]

I don't see any reason to assume a 45 degree angle or any angle other than straight up. One way to get an equation is to use "F= ma" or "a= F/m". For gravity, F= -mg so that just becomes a= -g. Then v= -gt+ v0 and x= -(g/2)t^2+ v0t= t(v0- gt). That's a parabola. Where is the vertex?

The other way is to use "conservation of energy". At 10m the total energy, potential energy plus kinetic energy, is, since the velocity is 0, is just the potential energy= mgh= 10mg. When it just starts, the potential energy is 0 so the kinetic energy is (1/2)mv^2= 10mg. What is v?

 

[Blobikins]

Ah, I'm sorry, I forgot to say a 8.99. M wall 40 m away, so the dx would be 80, sorry.What am I missing?

 

[CWatters]

If you assume 45 degree launch angle you do have enough knowns values to solve for the launch velocity. Look up the SUVAT equations..

 

[nasu]

If you assume the angle is 45 degree you have too many restrictions. The object won't be "just clearing the wall".
What you should assume is that the trajectory has its highest point at 10 m above the ground.

 

[Blobikins]

Alright, so I've finished the equation 1 way, with the peak being 10 m, but my teacher wants it done two ways. He suggested choosing 45 degrees as an angle, but how exactly would I go about doing that?

 

[nasu]

But 45 degrees does not satisfy the setup of the problem. How can you solve with contradictory conditions?
It's either "just clearing the wall" or 45 degrees angle. Given that you keep the two distances (10 m and 40 m) unchanged.
You need to have the problem very clearly formulated before thinking about solution.

 

[Windadct]

By my thinking 45 Deg (frictionless) exactly satisfies the set up - as the optimal, lowest launch Velocity - granted the problem does not state optimal !

 

[Blobikins]

So, I was thinking so at 40m it's just reaching 10m, not the peak at that point.

With the angle of 45 degrees, how do I do that?

 

[Windadct]

I am reading again and is the problem really "find two WAYS" -- Vs two solutions? ( there are infinite solutions)

 

[Blobikins]

Apologies, I meant 2 solutions.

Would it be possible for me to do a theoretical parabola where the midpoint is 80,20, so then a quarter of the time would be 40,10?

 

[Windadct]

It is a single math problem - and pick two solutions. The advantage of the 45 is since it is optimal it is the minimum V init.

 

[Blobikins]

I don't know what you're getting at.

Pick two solutions?

I feel I'm missing a variable if I just have 45?

What exactly do I do?

 

[Blobikins]

So basically I'td be x - ?cos45 = vx
d = 40

y =

?sin45 = viy
d =10
a = -9.8
t = ?

How do I put this together to make a proper equation

 

[CWatters]

If you assume 45 degree launch angle you do have enough knowns values to solve for the launch velocity. Look up the SUVAT equations..

The SUVAT equation I was hoping you would find is...

s = ut + 0.5at²

u = Vsin(45)
|a| = 9.8m/s/s
s = 9.99m

 

[insightful]

Brute-force method: Construct a parabola starting at (0,0) with a 45^o launch angle, peaking at, say, (B,A) and landing at (2B,0).
The general form is: y = A-C(x-B)^2.
You know another point on the curve: (40,10).
You know dy/dx at (0,0) = 1.
Solve for A, B, and C.