Hard projectile motion problem

In summary, the conversation discusses the standing broad jump and how it is performed by squatting and pushing off with the legs. It mentions that the extension of the legs from the crouch position is 0.600 m and the acceleration achieved is 1.25 times the acceleration due to gravity. The question is posed of how far one can jump, and it is mentioned that swinging the arms in the direction of the jump can increase the range. Assumptions are also mentioned, such as the possibility of swinging the arms in circles and the potential impact on distance depending on how the jumper lands.
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In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
 
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  • #2
Welcome to PF.
Is this homework?

Does the 1.25g acceleration take place vertically or maybe at 45°?
Arms forward on take-off will advance the centre of mass, but not necessarily give an increased distance unless they end up behind.
Swinging the arms in circles can change orientation while in the air. If the jumper was tall and landed flat on their back, lying across their track, they might go further. Do they have to land on their feet?
 

Related to Hard projectile motion problem

1. What is a hard projectile motion problem?

A hard projectile motion problem is a physics problem that involves calculating the motion of an object that is launched into the air with a known initial velocity and angle, and then falls under the influence of gravity. These problems are often considered "hard" because they require the use of advanced mathematical concepts and equations.

2. What is the formula for calculating the trajectory of a projectile?

The formula for calculating the trajectory of a projectile is y = y0 + xtanθ - (gx2 / 2v02cos2θ), where y is the vertical position, x is the horizontal position, y0 is the initial vertical position, v0 is the initial velocity, g is the acceleration due to gravity, and θ is the launch angle.

3. How do you find the maximum height of a projectile?

The maximum height of a projectile can be found by using the formula ymax = (v02sin2θ) / 2g, where v0 is the initial velocity and θ is the launch angle. This formula is derived from the fact that at the highest point of the trajectory, the vertical velocity is equal to zero.

4. What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = (v02sin2θ) / g, where v0 is the initial velocity and θ is the launch angle. This formula is derived from the fact that at the highest point of the trajectory, the vertical position is zero.

5. How do air resistance and other external factors affect projectile motion?

Air resistance and other external factors can affect projectile motion by changing the acceleration and velocity of the object. In most cases, air resistance will decrease the range and maximum height of a projectile. Other factors such as wind and air density can also have an impact on the trajectory of the projectile. These factors can be taken into account in more advanced projectile motion problems, but are often neglected in basic calculations.

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