A0.5 kg pendulum bob is raised 1.0 m above the floor. the bob is then released from rest. when the bob is 0.8 m above the floor, what is the speed

When the Bob is 0.8 m above the floor,the speed is; 1.98 m/s

We are given;

Mass of pendulum bob; m = 0.5 kg

Initial height; h1 = 1 m

Final height; h2 = 0.8 m

From conservation of energy, we know that;

PE1 + KE1 = PE2 + KE2

Formulas for potential and kinetic energy are;

KE = ½mv²

PE = mgh

Thus;

(0.5 × 9.8 × 1) + 0 = (0.5 × 9.8 × 0.8) + ½ × 0.5 × (v_2)²

Divide through by 0.5 to get;

9.8 = 7.84 + ½(v_2)²

(v_2)² = 2(9.8 - 7.84)

v_2 = 1.98 m/s

Read more about conservation of energy at; link

Here in this case we can use energy conservation law to find out the speed

As per energy conservation we know that

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

so the speed of the pendulum bob is approx 2 m/s when it is 0.8 m above the floor