A pears rolls across a table with a speed of 0,5 m/s and falls off the edge, if it lands 3m from the edge of the table from what height does it fall
I will assume that the "3 m from the edge of the table" is the horizontal distance, so
3 = .5t, where t is the time in seconds
t = 6 seconds
vertical distance = 4.9t^2 = 4.9(36) = 176.4 m
not a very logical answer.
So let's assume the 3 m is the direct line from the edge to where it lands, then
(.5t)^2 + (4.9t^2)^2 = 3
.25t^2 + (4.9t^2)^2 = 3
lt t^2 = x
.25x + 24.01x^2 = 3
2401x^2 + 25x - 300 = 0
x = (-25 ± √2881825)/4802
= .3483 or a negative x
t^2 = .3483 ---> t = .5902
height of table = 4.9(.5902)^2 = 1.71 metres
a more logical answer
To determine the height from which the pear falls, we can use the principles of kinematics and the equations of motion.
Assuming the motion of the pear is accelerated due to gravity, we can use the following kinematic equation:
s = ut + (1/2)gt^2
Where:
s = vertical distance or height of the fall
u = initial velocity (0 m/s)
t = time
g = acceleration due to gravity (approximately 9.8 m/s²)
In this scenario, the pear rolls with a horizontal velocity of 0.5 m/s until it falls off the edge of the table. The time it takes for the pear to hit the ground can be calculated using the horizontal distance traveled and the horizontal velocity:
time = distance / velocity
time = 3 m / 0.5 m/s
time = 6 seconds
Now that we have the time, we can substitute it back into the equation and solve for the vertical distance s:
s = (1/2)gt^2
s = (1/2)(9.8 m/s²)(6 s)^2
s = (1/2)(9.8 m/s²)(36 s²)
s = 176.4 m
Therefore, the pear falls from a height of approximately 176.4 meters.