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?
H = 12*t + (g/2)*t^2
Plug in t=2 and solve for the height H, which is the distance it falls in that time.
g = 9.8 m/s^2
To find the height of the tower, we can use the equations of motion. The equation we need to use is:
h = ut + (1/2)gt^2
Where:
h = height of the tower (what we need to find)
u = initial velocity of the rock (12 m/s, as given)
t = time taken for the rock to hit the ground (2.0 s, as given)
g = acceleration due to gravity (this is a constant, approximately 9.8 m/s^2)
Now, let's substitute the given values into the equation and calculate the height:
h = (12 m/s)(2.0 s) + (1/2)(9.8 m/s^2)(2.0 s)^2
Simplifying this equation:
h = 24 m + (1/2)(9.8 m/s^2)(4.0 s^2)
h = 24 m + 19.6 m
h = 43.6 m
Therefore, the height of the tower is 43.6 meters.