# Physics- Horizontal Displacement

Currently working my way through a maths problem, and I've hit a brick wall.

A circus acrobat is launched from a cannon that is aimed at 75 degrees above the ground. He leaves the barrel at a height of 4 meters, travelling at 6.5 meters per second. If the net must be placed at a height of 5.5 meters above the ground for safety, at what horizontal distance from the end of the cannon barrel should the centre of the net be placed so as to catch the acrobat exactly at that position? (Ignore air resistance, assume the acrobat's position can be approximated by a point and that g = 10 m/s^2. The centre of the net should be located ???? metres from the end of the cannon barrel.

Now, I know how to calculate the horizontal and vertical velocities, and I know the majority of the solving out for this, but I just cannot get the right answer. If anyone can show me the right way of solving this, I'd greatly appreciate it.

Thanks

• Welcome to Math.SE! If you can already solve some parts of the question, please include your own work so other people don't have to repeat it and can give you better feedback on what to do. Moreover, I think your question is better suited for Phys.SE, so I vote to migrate, even though the homework policy there is more stringent. – Hrodelbert May 8 '15 at 12:18
• Cheers for the welcome. As for migration, I'm not fussed where this goes (this isn't technically "homework"; at least, I'm not being assessed on this question, I'm doing it so that when the homework comes around I know how to solve the question.) – iskelz May 8 '15 at 12:23
• As far as I know, questions of this kind are always considered "homework" at Phys.SE, even though for you personally it is not. They just want to focus on more research oriented questions. – Hrodelbert May 8 '15 at 12:25
• Ah fair enough then – iskelz May 8 '15 at 12:29

This problem involes solving the differential equation $$(\dot{x}, \dot{y}) = (vx, vy)$$ and applying boundary conditions at the start and end of the flight to select the proper solution and infer the distance $d$. The physics is in providing the proper velocity $v(t)$.

Initial position and velocity (omitting units): $$(x_0, y_0) = (0, 4) \quad (vx_0, vy_0) = 6.5 \, (\cos 75^\circ, \sin 75^\circ)$$ Final position: $$(x_1, y_1) = (d, 5.5)$$ Motion: $$(x,y) = \int\limits_0^t (vx(\tau), vy(\tau)\, d\tau + (x_0, y_0)$$ If you got the velocity components, this should be easy.

What is left is to look for $t_1$, the time to reach $y_1$. This will then allow to calculate $d = x_1$.

I get $d \approx 1.572$.

With an initial velocity vector of $6.5$ meters/second at an angle of $75$ degrees from horizontal, what is the vertical component of the acrobat's initial velocity?

What is the horizontal component of the acrobat's velocity?

With $10$ meters/second/second acceleration downward, in how many seconds $t_1$ does the acrobat's vertical velocity become zero?

Given the initial height $4$ meters above the ground, at what height $h_1$ meters above the ground is the acrobat after $t_1$ seconds (when their vertical velocity reaches zero)?

How far does the acrobat travel horizontally in $t_1$ seconds?

How many seconds $t_2$ does it take to fall from $h_1$ meters to $5.5$ meters?

How far does the acrobat travel horizontally in $t_2$ seconds?

Now where should the net be?

It may help if you draw yourself a diagram showing each of these events and the path the acrobat follows between them.