A certain freely failing object requires 1.70 s to travel the last 20.5 m before it hits the ground. From what height above the ground did it fall?
Let H be the initial height.
Let T be the time it takes to fall the full height H.
g/2 T^2 = 4.9 T^2 = H
(g/2)(T-1.7)^2 = 4.9(T-1.7)^2 = H - 20.5
Combine the equations to eliminate H, so you can solve for T. Then calculate H.
4.9T^2 = 4.9(T-1.7)^2 + 20.5
4.9T^2 = 4.9T^2 -16.66 T +14.161 +20.5
16.66T = 34.66
T = 2.08 s
H = 4.9 T^2 = 21.2 m
Find the height from which you would have to drop a ball so that it would have a speed of 6.4 m/s just before it hits the ground.
To find the height from which the object fell, we can use the equation of motion:
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
In this case, we are given the time (t = 1.70 s) and the distance traveled (d = 20.5 m). However, we need to find the height, so we rearrange the equation to solve for h:
h = d - (1/2) * g * t^2
Now we can substitute the given values into the equation:
h = 20.5 m - (1/2) * 9.8 m/s^2 * (1.70 s)^2
h = 20.5 m - (1/2) * 9.8 m/s^2 * (2.89 s^2)
h = 20.5 m - 14.222 m
h = 6.278 m
Therefore, the object fell from a height of approximately 6.278 meters above the ground.