a very strong young child tosses a stone into the air. his father times the stone's travel and finds that it takes the stone 2.8 seconds to return to the child's hand. how high did the stone rise?
Y=(1/2)*(gravity)*(time)^2
where time=1.4 sec
takes 1/2 the time to go up and 1/2 the time to go down
thank you :)
To find the height the stone reached, we can use the concept of free fall and the equations of motion.
First, let's break down the stone's motion into two parts: upward and downward.
During the upward motion, the stone is accelerating against gravity, so we can use the equation:
h = ut + (1/2)at^2
Here,
h is the height,
u is the initial velocity (which is 0 in this case since the stone starts from rest),
t is the time taken for the upward journey, and
a is the acceleration due to gravity (-9.8 m/s^2).
Substituting the values, we get:
h = 0 + (1/2)(-9.8)(t/2)
= (-4.9)(t^2)
Since the upward journey and downward journey take equal amounts of time, we can write:
2(t/2) = 2.8
Simplifying, we find:
t = 2.8 seconds
Now we can substitute this value back into the equation for h:
h = (-4.9)(2.8^2)
= -4.9(7.84)
= -38.416
Since height cannot be negative, we take the absolute value of h:
h = 38.416 meters (rounded to three decimal places)
Therefore, the stone reached a height of approximately 38.416 meters.