A daredevil is shot out of a cannon

• MHB
• karush
In summary: Delta y = v_0 \tan{\theta_0} - \dfrac{g \cdot (\Delta x)^2}{2v_0^2 \sin^2{\theta_0}}$Substitute that result into the first equation to find$h$...$h = \dfrac{v_0 \tan{\theta_0}}{2g \cdot (\Delta x)^2}$karush Gold Member MHB$\textsf{ A daredevil is shot out of a cannon at $45^\circ$ to the horizontal with an initial speed of 25.0 m/s.}\textsf{ A net is positioned a horizontal distance of 50.0 m from the canon.}\\\textit{At what height above the cannon should the net be placed in order to catch the daredevil?}so far did this \begin{align*}\displaystyle v&=(25)\cdot\cos{45^\circ}=17.68 \end{align*} And "what you have done" makes no sense because you haven't said what "v" means! And it can't be "velocity" because the velocity here is a vector quantity. Here is what I would do- this is "acceleration due to gravity so the acceleration vector is <0, -g>. The initial velocity vector is $$\displaystyle \left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}\right>$$ (that "$$\displaystyle \frac{\sqrt{2}}{2}$$" is your "[FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]45[/FONT][FONT=MathJax_Main][/FONT]) so the velocity vector is $$\displaystyle \left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}- gt\right>$$ and, taking the initial position to be (0, 0), the position vector is $$\displaystyle \left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>$$. If the initial height of the cannon is "h" meters above the net, and the net is 50 meters from the cannon, then the net's position is <50, -h> so we must have $$\displaystyle \left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>= <50, -h>$$. That is the same as the two equations $$\displaystyle \frac{25\sqrt{2}}{2}t= 50$$ and $$\displaystyle \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2= -h$$. Solve the first equation for t then put that value of t into the second equation to find h.\Delta x = v_0 \cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0 \cos{\theta_0}}\Delta y = v_0 \sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2$substitute the expression for$t$derived from the first equation into the second ...$\Delta y = \Delta x \cdot \tan{\theta_0} - \dfrac{g \cdot (\Delta x)^2}{2v_0^2 \cos^2{\theta_0}}$you were given values for$\Delta x$,$v_0$, and$\theta_0$... calculate the value of$\Delta y\$.

What is the science behind a daredevil being shot out of a cannon?

The science behind a daredevil being shot out of a cannon involves principles of physics, specifically projectile motion and Newton's laws of motion. The cannon serves as a powerful force to launch the daredevil into the air, and the angle, speed, and trajectory of the launch must be carefully calculated to ensure a successful and safe landing.

How fast is a daredevil shot out of a cannon?

The speed at which a daredevil is shot out of a cannon can vary depending on factors such as the type and power of the cannon, the angle and trajectory of the launch, and the weight and size of the daredevil. However, on average, daredevils can reach speeds of 60-70 miles per hour when shot out of a cannon.

What safety precautions are taken when shooting a daredevil out of a cannon?

Safety is of the utmost importance when it comes to shooting a daredevil out of a cannon. Before the launch, the daredevil and crew must carefully inspect the cannon and ensure it is in good working condition. Protective gear, such as a helmet and padding, is also worn by the daredevil. Additionally, there are safety measures in place to control the angle and speed of the launch to minimize the risk of injury.

How far can a daredevil be shot out of a cannon?

The distance a daredevil can be shot out of a cannon depends on various factors, such as the power of the cannon, the angle and trajectory of the launch, and the weight and size of the daredevil. On average, daredevils can be launched anywhere from 50 to 100 feet in the air.

What training is required to be a daredevil shot out of a cannon?

Becoming a daredevil who is shot out of a cannon requires rigorous training and years of experience. Daredevils must have a strong understanding of physics and be physically fit to withstand the forces of the launch. They also undergo extensive training to learn how to safely land after being shot out of the cannon.

Similar threads

• General Math
Replies
1
Views
3K
• General Math
Replies
2
Views
2K
• General Math
Replies
4
Views
2K
• General Math
Replies
2
Views
6K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Calculus
Replies
7
Views
2K
• Calculus
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
403