A rock is thrown downward from an unknown height above the ground with an initial speed of 10m/s. It strikes the ground 3s later. Determine the initial height of the rock above the ground.
To determine the initial height of the rock above the ground, we can use the equations of motion. The equation that relates the height (h), initial velocity (v₀), time (t), and acceleration due to gravity (g) is:
h = v₀t + (1/2)gt²
In this case, the rock is thrown downward, so the initial velocity is negative (-10 m/s), the acceleration due to gravity is also negative (-9.8 m/s²), and the time is given as 3 seconds.
Plugging in these values into the equation, we can solve for the initial height (h):
h = (-10 m/s)(3 s) + (1/2)(-9.8 m/s²)(3 s)²
h = -30 m + (-4.9 m/s²)(9 s²)
h = -30 m - 44.1 m
h = -74.1 m
The negative sign indicates that the initial height is below the ground level. Therefore, the initial height of the rock above the ground is 74.1 meters below the ground.
hf=Ho-vo*t-4.9t^2
vo= -10m/s
hf=0
solve for Ho