An object is released from rest on a planet that has no atmosphere. The object falls freely for 5.02 m in the first second. What is the magnitude of the acceleration due to gravity on the planet?
Answer in units of m/s
can someone help me please? i tried doing this
-5.02m =Xa
-5.02m =Vo*(t) + (1/2)*a(t^2)
-5.02m =(1/2)*a
-10.04m/s^2 = a
but its says that its wrong. can someone please help?
Sure! Let's go through the problem step by step to figure out where the mistake might be.
The equation you used is correct:
d = V₀t + (1/2)at²
In this case, the initial velocity V₀ is 0 since the object is released from rest. The distance fallen, d, is given as 5.02 m, and the time, t, is given as 1 s.
Plugging in the values, we have:
5.02 m = (1/2)a(1 s)²
Now let's solve for the acceleration, a:
5.02 m = (1/2)a(1 s)²
5.02 m = (1/2)a(1 s)²
10.04 m = a(1 s)²
10.04 m/s² = a(1)²
From this, we can see that your calculations were correct, and the magnitude of the acceleration due to gravity on the planet is indeed 10.04 m/s².
I'm not sure why your answer was marked wrong, but please double-check your input and try again. If the problem persists, it might be worth discussing the issue with your teacher or professor.
I hope this helps! Let me know if you have any further questions.