Physics- Horizontal Displacement

Question

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

Answer 1

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 $d \approx 1.572$.

Answer 2

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.