An object of mass 0.7 kg is thrown vertically upwards at a speed of 23 m/s. What is the gravitational potential energy of the object at the highest point of its trajectory?
The GPE at the highest point will equal to the KE at launch: 1/2 m v^2
To find the gravitational potential energy of the object at the highest point of its trajectory, we need to determine the height at that point.
The initial vertical velocity of the object is 23 m/s. When the object reaches the highest point, its vertical velocity becomes zero before it starts falling back downwards due to gravity.
Using the kinematic equation for vertical motion, we can find the height (h) reached by the object:
v_f^2 = v_i^2 + 2aΔd
Since the final velocity (v_f) is 0 m/s and the initial velocity (v_i) is 23 m/s, we can rearrange the equation to solve for the displacement (Δd), which represents the height:
0 = (23 m/s)^2 + 2*(-9.8 m/s^2)*Δd
Simplifying the equation, we have:
-2 * 9.8 m/s^2 * Δd = (23 m/s)^2
Solving for Δd:
Δd = [(23 m/s)^2] / (2 * 9.8 m/s^2)
Δd ≈ 27.859 m
Now that we know the height reached by the object is approximately 27.859 meters, we can calculate the gravitational potential energy (PE) using the formula:
PE = m * g * h
Where m is the mass of the object (0.7 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (27.859 m):
PE = (0.7 kg) * (9.8 m/s^2) * (27.859 m)
PE ≈ 184.428 J
Therefore, the gravitational potential energy of the object at the highest point of its trajectory is approximately 184.428 Joules.