Velocity leaving a cannon, hitting landing cushion, and acceleration experienced

[pech0706]

Homework Statement

An 80 kg stuntman is shot from a cannon at a circus. The cannon is 3m in length and angled at 35 degrees above the level floor. The landing cusion is 30m away from the cannon. What velocity will the stuntman have 1)when he leaves the cannon and 2)when he first hits the landing cushion? What acceleration must the stuntman experience in the shot from the cannon? Ignore air resistance.

Homework Equations

V_o=$$\sqrt{dg/sin2\Theta}$$
V_o²=2a$$\Delta$$x

The Attempt at a Solution

My solutions are higher then the answers given. When i calculated initial velocity (for being shot out of the cannon?) i got 17.7 m/s. that was using d=30m, g=9.81 divided by sin 2(35). Using that number i got the acceleration to be 5.22m/s, for that one i believe the answer is around 52 m/s. I don't know exactly how to get the velocity at the landing cushion.

Any help is much appreciated!

 

[SammyS]

The cannon barrel is only 3.0 meters long.

You used 30 meters to find the accelerstion.

 

[pech0706]

alright, using 3m as the range for the range equation i got an initial velocity of 5.8m/s, but that got me an even smaller acceleration.

 

[SammyS]

Assuming that v₀ = 17.7 m/s is correct (and it appears that it is) then to find the acceleration experienced by the stunt man while he id in the cannon, use:

v₀² = 2 a_cannon Δx_cannon.

For this part, use 3.0 meters. (You apparently used 30 meters.)

 

[rwisz]

I would like to thank you for providing your attempt at a solution and the relevant equations. Your calculations seem to be correct, and the discrepancy between your answers and the given answers could be due to rounding errors. It is always important to double-check your calculations and make sure that you are using appropriate units.

To calculate the velocity at the landing cushion, we can use the equation for projectile motion:

Vf^2 = Vi^2 + 2ad

Where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the distance traveled. We know the initial velocity (17.7 m/s) and the distance traveled (30 m). We can assume that the final velocity is 0 m/s since the stuntman comes to a stop at the landing cushion. Therefore, we can rearrange the equation to solve for acceleration:

a = (Vf^2 - Vi^2)/2d

Plugging in the values, we get:

a = (0 - (17.7)^2)/2(30) = -7.41 m/s^2

Since the acceleration is negative, it means that the stuntman is slowing down as he approaches the landing cushion. This is expected since he is experiencing a deceleration due to the cushion's impact.

I hope this helps clarify any confusion and provides a more accurate solution to the problem. Keep up the good work!