Omar throws a rock down with speed 12 m/s from the top of a tower. The rock hits the ground after 2.0 s. What is the height of the tower? (air resistance is negligible)

To find the height of the tower, we can use the equation of motion:

\(h = v_0t + \frac{1}{2}gt^2\)

where:
- \(h\) is the height of the tower
- \(v_0\) is the initial velocity (rock thrown downward) which is -12 m/s (negative because it's directed downward)
- \(t\) is the time taken for the rock to hit the ground, which is 2.0 s
- \(g\) is the acceleration due to gravity, which is approximately -9.8 m/s² (negative because it's directed downward)

Substituting the given values into the equation, we can calculate the height of the tower:

\(h = (-12 \, \text{m/s})(2.0 \, \text{s}) + \frac{1}{2}(-9.8 \, \text{m/s²})(2.0 \, \text{s})^2\)

Simplifying this equation will give us the height of the tower.

hf=hi+vi*t+1/2 g t

hf=0, Vi=-12, g= -9.8m/s^2 solve for t.