Q: a freely falling object requires 1.50s to travel the last 30.0m before it hits the ground . From what height above the ground did it fall ??
Pleazzzz help ....
Let H be the height from which it is dropped. Let T be the time it takes to hit the ground. Solve these two simultaneous equations:
H = (1/2) g T^2
H-30 = (1/2) g (T-1.5)^2
By subtracting equations, H can be eliminated immediately:
30 = 1.5 g T - (1/2) g (1.5)^2
Solve for T first, then compute H.
t= 2.79 H=38.1
To determine the height from which the object fell, we can use the equations of motion for a freely falling object. The equation we will use is:
h = (1/2) * g * t^2
Where:
h is the height
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken for the object to fall (given as 1.50 s).
Plugging in the given values into the equation:
h = (1/2) * (9.8 m/s^2) * (1.50 s)^2
h = 7.35 m
Therefore, the object fell from a height of 7.35 meters above the ground.
To find the height from which the object fell, we can make use of the equations of motion for freely falling objects.
First, let's define some variables:
- h = height from which the object fell (what we need to find)
- t = time taken to travel the last 30.0m = 1.50s
- g = acceleration due to gravity = 9.8 m/s^2 (approximately, at the surface of the Earth)
Using the equation of motion for vertical displacement:
h = (1/2) * g * t^2
Plugging in the values we have:
h = (1/2) * (9.8 m/s^2) * (1.50s)^2
h ≈ 1/2 * 9.8 m/s^2 * 2.25s^2
h ≈ 1/2 * 9.8 m/s^2 * 5.0625 m^2/s^2
h ≈ 49.05 m^2/s^2
So, the object fell from a height of approximately 49.05 meters above the ground.