A body of mass 5Kg, thrown up Vertically with a velocity of 9-8 m/s. What is the KE at the mount propetu - Os and 0.5s and 1send

The initial velocity $u = 9.8 \text{ m/s}$.
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.

$$KE_{0 \text{ s}} = \frac{1}{2} (5 \text{ kg}) (9.8 \text{ m/s})^2$$ $$KE_{0 \text{ s}} = \frac{1}{2} (5 \text{ kg}) (96.04 \text{ m}^2/\text{s}^2)$$ $$KE_{0 \text{ s}} = 240.1 \text{ J}$$

Using the kinematic equation $v = u - gt$, where $g = 9.8 \text{ m/s}^2$.

$$v_{0.5 \text{ s}} = 9.8 \text{ m/s} - (9.8 \text{ m/s}^2)(0.5 \text{ s})$$ $$v_{0.5 \text{ s}} = 9.8 \text{ m/s} - 4.9 \text{ m/s}$$ $$v_{0.5 \text{ s}} = 4.9 \text{ m/s}$$

$$KE_{0.5 \text{ s}} = \frac{1}{2} (5 \text{ kg}) (4.9 \text{ m/s})^2$$ $$KE_{0.5 \text{ s}} = \frac{1}{2} (5 \text{ kg}) (24.01 \text{ m}^2/\text{s}^2)$$ $$KE_{0.5 \text{ s}} = 60.025 \text{ J}$$

Using the kinematic equation $v = u - gt$.

$$v_{1 \text{ s}} = 9.8 \text{ m/s} - (9.8 \text{ m/s}^2)(1 \text{ s})$$ $$v_{1 \text{ s}} = 9.8 \text{ m/s} - 9.8 \text{ m/s}$$ $$v_{1 \text{ s}} = 0 \text{ m/s}$$

$$KE_{1 \text{ s}} = \frac{1}{2} (5 \text{ kg}) (0 \text{ m/s})^2$$ $$KE_{1 \text{ s}} = 0 \text{ J}$$

The kinetic energies are:
At $t = 0 \text{ s}$: $\boxed{\text{240.1 J}}$
At $t = 0.5 \text{ s}$: $\boxed{\text{60.025 J}}$
At $t = 1 \text{ s}$: $\boxed{\text{0 J}}$