Two pole-vaulters just clear the bar at the same height. The first lands at a speed of 8.50 m/s, and the second lands at a speed of 8.80 m/s. The first vaulter clears the bar at a speed of 1.50 m/s. Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

landing speed =starting speed

v₀₁ = 8.5 m/s, v₀₂ = 8.8 m/s,
v₁ = 1.5 m/s, v₂ = ?

h=(v₁²-v₀₁²)/2g,
h=(v₂²-v₀₂²)/2g,
(v₁²-v₀₁²)/2g=(v₂²-v₀₂²)/2g,
v₂=sqrt(v₁²+v₀₂²-v₀₁²) =
= sqrt(1.5²+8.8²-8. 5²)= 2.73 m/s.

2.73

2.73

Well, it seems like these pole-vaulters are really on top of their game! As for the second vaulter's speed when clearing the bar, let's just hope they don't get too speedy and vault themselves into the land of uncontrollable speed. The first vaulter cleared the bar at a whopping 1.50 m/s, which is quite impressive. And since both pole-vaulters cleared the bar at the same height, we can assume that they are equally skilled in the art of pole-vaulting. So, if the first vaulter landed at 8.50 m/s and the second vaulter landed at 8.80 m/s, the second vaulter's speed when clearing the bar must be somewhere in between. Perhaps they jumped with the grace of a ballet dancer, or maybe they were just really eager to impress the judges. Either way, it's safe to assume that the second vaulter cleared the bar at a speed between 1.50 m/s and 8.80 m/s. Hope that sheds some light on the situation!

To find the speed at which the second vaulter clears the bar, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (the vaulter and the Earth) is conserved in the absence of external forces such as air resistance and friction.

The mechanical energy of an object can be defined as the sum of its kinetic energy (KE) and potential energy (PE). In this case, the potential energy is due to the height gained during the vault.

We can write the conservation of mechanical energy equation as:

Initial KE1 + Initial PE1 = Final KE1 + Final PE1,
Initial KE2 + Initial PE2 = Final KE2 + Final PE2.

Given that the potential energy at the peak of the vault is zero for both vaulters, the equations simplify to:

Initial KE1 = Final KE1,
Initial KE2 = Final KE2.

The initial kinetic energy (Initial KE) is given by:

Initial KE = (1/2) * mass * velocity^2.

Let's assume that both vaulters have the same mass.

For the first vaulter:
Initial KE1 = (1/2) * m * (8.50 m/s)^2 = 36.125 m^2/s^2.

For the second vaulter:
Initial KE2 = (1/2) * m * (8.80 m/s)^2 = 40.144 m^2/s^2.

Since the initial kinetic energy is conserved, the speed at which the second vaulter clears the bar is the square root of the ratio of the initial kinetic energies:

v2 = sqrt(Initial KE2 / Initial KE1) = sqrt(40.144 m^2/s^2 / 36.125 m^2/s^2) = 1.07 m/s.

Therefore, the speed at which the second vaulter clears the bar is approximately 1.07 m/s.