A pole-vaulter of mass 60.5 kg vaults to a height of 5.8 m before dropping to thick padding placed below to cushion her fall.(a) Find the speed with which she lands.

To solve this problem, we can use the principle of conservation of energy. This principle states that the total energy of an isolated system remains constant if no external forces act on it. In this case, the pole-vaulter's total energy at the peak of her vault (potential energy) will be equal to her total energy just before she lands (kinetic energy).

Step 1: Calculate the potential energy at the peak of the vault. Potential energy (PE) is given by the formula PE = mgh, where m is mass, g is the acceleration due to gravity, and h is height.

PE = (60.5 kg)(9.8 m/s^2)(5.8 m) = 3427.06 Joules

Step 2: Set the potential energy equal to the kinetic energy just before landing. Kinetic energy (KE) is given by the formula KE = 1/2 mv^2, where m is mass and v is velocity (the speed we're trying to find).

3427.06 J = 1/2 (60.5 kg) v^2

Step 3: Solve for v. First, multiply both sides of the equation by 2 to get rid of the 1/2 on the right side:

6854.12 J = (60.5 kg) v^2

Then, divide both sides by 60.5 kg to isolate v^2:

v^2 = 6854.12 J / 60.5 kg = 113.27 (m^2/s^2)

Finally, take the square root of both sides to solve for v:

v = sqrt(113.27 m^2/s^2) = 10.64 m/s

So, the pole-vaulter lands with a speed of approximately 10.64 m/s.