when an object is dropped frome a hight of 20m above the surface of planet Z, it will fall 5m during the 2nd second of fall. What is the acceleration of a falling object near the surface of this planet?
h(2)-h(1)=5 given
1/2 g*2^2 - 1/2 g*1^2=5
g(2-.5)=5 you do the rest.
To determine the acceleration of a falling object near the surface of planet Z, we can use the kinematic equation for the distance fallen by an object during free fall:
d = (1/2) * a * t^2
where:
d is the distance fallen (5m),
a is the acceleration, and
t is the time (2 seconds).
Plugging in the given values, we have:
5 = (1/2) * a * 2^2
Simplifying the equation, we have:
5 = (1/2) * 4a
Now, we solve for 'a' by isolating it:
4a = 5 * 2
4a = 10
a = 10/4
a = 2.5 m/s^2
Therefore, the acceleration of a falling object near the surface of planet Z is 2.5 m/s^2.
To find the acceleration of a falling object near the surface of planet Z, we can use the equations of motion for free fall. The equation we will use is:
d = (1/2) * a * t^2
where:
- d is the distance fallen by the object
- a is the acceleration
- t is the time of fall
Given that the object falls 5m during the 2nd second of fall, we can substitute these values into the equation:
5m = (1/2) * a * (2s)^2
Simplifying the equation:
5m = (1/2) * a * 4s^2
Divide both sides of the equation by 4 to isolate the acceleration:
a = (5m) / ((1/2) * 4s^2)
a = (5m) / (2s^2)
a = 2.5 m/s^2
Therefore, the acceleration of a falling object near the surface of planet Z is 2.5 m/s^2.