given a time of fall t = 1.07 s
what is the initial height h? You may assume that the acceleration due to gravity
g =9.792 𝑚/s^2
(1.07)^2 x 9.792 +11.21 m is this correct?
s = 1/2 a t^2
4.896 * 1.07^2 = 5.605 meters
where did you get the 11.21?
ahhh. you mean = instead of +, but forgot the factor of 1/2
I have been trying to wrap my head but why is there a 1/2
To find the initial height, we can use the equation of motion for free fall:
h = (1/2)gt^2 + vt + h0
Where:
h = final height (initial height + distance fallen)
g = acceleration due to gravity (9.792 m/s^2)
t = time of fall (1.07 s)
v = final velocity (0 m/s, since the object has reached the ground)
h0 = initial height (what we're trying to find)
Since the object has reached the ground, the final velocity (v) is zero. Therefore, the equation simplifies to:
h = (1/2)gt^2 + h0
Plugging in the given values:
h = (1/2)(9.792 m/s^2)(1.07 s)^2 + h0
h = 5.227 m + h0
So, the correct equation should be h = 5.227 m + h0.
To find the initial height, we need more information.