An object of mass 100g is thrown vertically upward from a point 60cm upabove the earth's surface with an initial velocity of 150cm/sec
well, the height in meters is
h(t) = 0.60 + 0.15t - 4.9t^2
I presume you want to do something with its weight, which is 0.1*9.8 N
. . .
To find the maximum height reached by the object, we can use the equations of motion. The equation we will use is:
v^2 = u^2 + 2aS
where:
v = final velocity (0 m/s when the object reaches the maximum height)
u = initial velocity (150 cm/s)
a = acceleration (due to gravity, which is approximately 9.8 m/s^2)
S = displacement (maximum height reached by the object)
First, let's convert the given values into SI units:
Mass of the object = 100g = 0.1 kg
Initial velocity = 150 cm/s = 1.5 m/s
Acceleration due to gravity = 9.8 m/s^2
Displacement = 60 cm = 0.6 m
Now, let's calculate the maximum height reached:
0 = (1.5 m/s)^2 + 2 * (-9.8 m/s^2) * S
Rearranging the equation:
0.9 m^2/s^2 - 19.6 m/s^2 * S = 0
Simplifying further:
S = (0.9 m^2/s^2) / (19.6 m/s^2)
S ≈ 0.046 m
Therefore, the maximum height reached by the object is approximately 0.046 meters (or 4.6 cm) above the initial point.