Speed at which to hurl a projectile?

[vabtzeexo]

Homework Statement

The Archimedes' catapult could hurl a 77 kg stone a distance of 180 m. What was the stone's initial velocity as it left the catapult? The wind resistance is assumed to be insignificant.

Homework Equations

$$y = v_y0 t - (1/2) g t^2$$
$$x = v_x0 t$$

The Attempt at a Solution

I think the mass of the stone is irrelevant as wind resistance is not taken into account. The problem would be easy if I knew the angle at which the catapult was fired. Here, however, both the angle and the initial velocity of the projectile are unknown. The answer provided by the textbook says the velocity is > 42 m/s. So clearly the 42 m/s is the initial velocity at the optimal angle and if you change the angle, you have to make up for it by launching at a greater initial velocity. I think the solution might have something to do with a derivative function, but other than that, I'm completely lost.

 

[Redbelly98]

Welcome to Physics Forums.

There is an optimal angle, which gives the maximum range for a projectile launched from ground level. Perhaps your book has a discussion of "range" in the section on projectile motion?

 

[vabtzeexo]

Thanks. I figured it out now. So apparently the optimal angle is always 45 degrees.

 

[Redbelly98]

Yes. Glad it worked out.

 

[rwisz]

I would first clarify the assumptions and variables of the problem. It is stated that the wind resistance is insignificant, so it can be assumed that the only force acting on the projectile is gravity. The mass of the stone is also given, but as mentioned, it is irrelevant in this case.

To determine the initial velocity of the projectile, we can use the equations of projectile motion. The vertical equation is given by y = v_y0 t - (1/2) g t^2, where v_y0 is the initial velocity in the y-direction (vertical) and g is the acceleration due to gravity. Since the stone is being launched horizontally, the initial velocity in the y-direction is 0. This means that the equation can be simplified to y = -(1/2) g t^2.

The horizontal equation is given by x = v_x0 t, where v_x0 is the initial velocity in the x-direction (horizontal). We can rearrange this equation to v_x0 = x/t.

Plugging in the given values of distance (180 m) and time (unknown), we can solve for the initial horizontal velocity. However, we still need to determine the time of flight.

To find the time of flight, we can use the fact that the projectile reaches its maximum height at the halfway point of its flight. This means that the time of flight is half of the total flight time. Using the vertical equation, we can set y = 0 and solve for t. This gives us t = √(2y/g).

Substituting this value for t into the horizontal equation, we get v_x0 = x/√(2y/g). Plugging in the given values, we get v_x0 = (180 m)/√(2(0.5)(9.8 m/s^2)), which simplifies to v_x0 = 18.41 m/s.

Therefore, the initial velocity of the projectile is 18.41 m/s in the horizontal direction. However, this is not the optimal angle for maximum distance. As stated, the textbook answer is >42 m/s, which means that the initial velocity would have to be increased to achieve this distance. This could be done by increasing the angle at which the catapult is fired, as you mentioned.

In conclusion, the initial velocity of the projectile can be calculated using the equations of projectile motion, but the optimal angle for maximum distance