a) State the principle of conservation of energy - Leaving Cert Physics - Question Question 1 - 2008

The graph consists of two parts:

1. From t = 0 to t = 3.0 seconds, the velocity increases from 0 to 9.2 m/s, representing the athlete's acceleration.
2. From t = 3.0 seconds to t = 5.0 seconds, the velocity remains constant at 9.2 m/s, indicating the athlete's maintained speed before the jump. Both axes should be labeled: the x-axis for time (seconds) and the y-axis for velocity (m/s).

To calculate the distance travelled before jumping, we can use the formula:

ext{Distance} = rac{1}{2} a t^2 + v t \

Given the first stage:

• From rest (u = 0), acceleration can be calculated as:
• Acceleration (a) = change in velocity / time = (9.2 m/s) / (3.0 s) = 3.0667 m/s²
• Distance during acceleration:

ext{Distance}_{ ext{acc}} = rac{1}{2} imes 3.0667 imes (3.0)^2 = 13.8 ext{ m}

During the constant speed phase (2.0 seconds):

$$ext{Distance}_{ ext{const}} = 9.2 ext{ m/s} imes 2.0 ext{ s} = 18.4 ext{ m}$$

Total distance travelled:

$$ext{Total Distance} = 13.8 + 18.4 = 32.2 ext{ m}$$

To find the maximum height the athlete can reach, we first convert the kinetic energy into potential energy. The kinetic energy (K.E.) at maximum velocity is:

ext{K.E.} = rac{1}{2} mv^2 \

Assuming the mass cancels out, we use:

• Maximum velocity (v) = 9.2 m/s

ext{K.E.} = rac{1}{2} imes 9.2^2 = 42.32 ext{ J} \

Setting this equal to potential energy at maximum height:

ext{P.E.} = mgh \

Using the height formula:

• Gravitational acceleration (g) = 9.8 m/s²,

Solving:

ext{Max Height} = rac{42.32}{9.8} \

The height the athlete can raise his centre of gravity from the ground is:

$$$$

Thus, the maximum height above the ground is 5.42 m.