Question

a 710 n daredevil was shot out of a cannon at an upward angle. he landed 2.3 s later in a net at the same height but 35m away from the cannon.

A)what was the magnitude of his initial velocity?
B)what was the launch angle?
C)it took the cannon 0.6 seconds to launch him and the kinetic friction in the cannon was 405 n. what constant force did the cannon apply to the daredevil? (hint after the second attempt.)

Answer 1

Solution

We have weight of man in surface od Earth $P = 710N$, from hence mass of man is $m = 72.38kg$.

We have an equation of motion on $x$- and $y$-axis's (launch angle is $\alpha, \nu_0$ is magnitude of initial speed)

$$x = \nu_0 t \cos \alpha$$ $$y = \nu_0 t \sin \alpha - g \frac{t^2}{2}$$

From hence we get about flight time $t = \frac{2\nu_0\sin\alpha}{g} = 2.3s$, distance of flight

$$I = \frac{2\nu_0^2 \cos \alpha \sin \alpha}{g} = 35m.$$

From hence, we get launch angle $\tan \alpha = \frac{gt^2}{2I} = 0.741$

From hence we get $\nu_0 = \frac{gt}{2\sin\alpha} = \frac{gt}{2}\sqrt{1 + \tan^{-2}\alpha} = 18.92m/s$

We have that the in canon

$$m\nu_0 = \Delta t (F - F_r - mg \sin \alpha)$$ $$F_r = 405N$$ $$\Delta t = 0.6s,$$ $$mg \sin \alpha = 422.7N$$ $$F = 3110N$$

Answer:

A) $\nu_0 = 18.92m/s$

B) $\tan \alpha = \frac{gt^2}{2I} = 0.741$

C) $F = 3110N$