An apple (0.100 kg) is thrown vertically upward at a speed of [let x = any whole number between 10 and 20] m/s (from an arm 1.5 m off of the ground). What is the maximum height that the apple reaches? (g=9.8 m/s2)
To find the maximum height reached by the apple, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (let's say x m/s)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement or change in height
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (0 - x^2) / (2 * (-9.8))
Simplifying further:
s = -x^2 / 19.6
Let's calculate the maximum height reached by substituting a whole number value of x between 10 and 20 into the equation.
To find the maximum height that the apple reaches, we can use the equations of motion.
First, we need to determine the initial velocity of the apple. It is given that the apple is thrown vertically upward with a speed of x m/s. Since x can be any whole number between 10 and 20, we can substitute the value of x into the equation.
Let's say x = 15 m/s. Therefore, the initial velocity of the apple (u) is 15 m/s.
Next, we need to find the time it takes for the apple to reach its maximum height. To do this, we can use the equation:
v = u + at,
where v is the final velocity (which is 0 m/s at maximum height) and a is the acceleration due to gravity.
Rearranging the equation, we have:
0 = 15 - 9.8t.
Solving for t, we get:
9.8t = 15,
t = 15 / 9.8,
t ≈ 1.53 s.
So, it takes approximately 1.53 seconds for the apple to reach its maximum height.
Now, to find the maximum height (h), we can use the equation:
h = ut + (1/2)at^2.
Substituting the values:
h = 15 × 1.53 + (1/2) × 9.8 × (1.53)^2,
h ≈ 11.54 m.
Therefore, the maximum height that the apple reaches is approximately 11.54 m.
Vf^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g + 1.5 = Meters above gnd.
Vo = 20 m/s.
g = -9.8 m/s^2.