omar throws a rock down with speed 12 m/s from the top of a tower. the rock hits the ground after 2 s. what is the height of the tower?
use this formula,
d=v1(t)+.5(a)(t^2)
So, 12(2)+.5(9.8)(2^2)
Hope that helps.
To find the height of the tower, we can use the equation of motion:
h = ut + (1/2)gt^2
where:
h is the height of the tower,
u is the initial velocity,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time taken.
Given:
u = 12 m/s
t = 2 s
Plugging these values into the equation, we have:
h = (12 m/s)(2 s) + (1/2)(9.8 m/s^2)(2 s)^2
Simplifying the equation further:
h = 24 m + (1/2)(9.8 m/s^2)(4 s^2)
h = 24 m + (1/2)(9.8 m/s^2)(16 s^2)
h = 24 m + (1/2)(9.8 m/s^2)(16 s^2)
h = 24 m + (1/2)(9.8 m/s^2)(16 s^2)
h = 24 m + (1/2)(9.8 m/s^2)(16 s^2)
h = 24 m + 78.4 m
h = 102.4 m
Therefore, the height of the tower is 102.4 meters.
To find the height of the tower, we can use the formula for free fall motion:
d = (1/2) * g * t^2
where,
d = distance or height
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
In this case, the rock falls for 2 seconds, so we can substitute those values into the formula:
d = (1/2) * 9.8 * (2^2)
d = 4.9 * 4
d = 19.6 meters
Therefore, the height of the tower is 19.6 meters.