a rock was thrown at the height of 50.0 m above the ground with an initial velocity of 20.0 m/s straight upward.
I assume it came down again.
Do you have a question?
To find certain information regarding the rock, we can use the equations of motion and the laws of physics. Let's break down the problem step by step.
1. Initial height: The rock was thrown at a height of 50.0 m above the ground. This means that the initial height (H) of the rock above the ground is 50.0 m.
2. Initial velocity: The rock was thrown straight upward with an initial velocity (V0) of 20.0 m/s. This means that the rock was moving upward with a speed of 20.0 m/s.
3. Acceleration due to gravity: When an object is thrown upward near the surface of the Earth, it experiences an acceleration due to gravity that acts in the opposite direction of its motion. The acceleration due to gravity (g) is approximately -9.8 m/s^2 (negative sign indicates that it acts downward).
4. Time taken to reach maximum height: To calculate the time(t) taken for the rock to reach maximum height, we use the formula:
V_final = V_initial + (acceleration × time)
0 m/s = 20.0 m/s - (9.8 m/s^2 × t)
Solving this equation for time(t), we find t = 20.0 m/s ÷ 9.8 m/s^2 = 2.04 s
5. Maximum height reached: To calculate the maximum height(H_max) reached by the rock, we use the kinematic equation:
H_max = H_initial + (V_initial × t) + (0.5 × acceleration × t^2)
H_max = 50.0 m + (20.0 m/s × 2.04 s) + (0.5 × -9.8 m/s^2 × (2.04 s)^2)
H_max = 50.0 m + 40.8 m - 20.06 m
H_max = 70.74 m
Therefore, the maximum height reached by the rock is approximately 70.74 meters above the ground.