The ceiling of a classroom is 3.75 m above the floor. A student tosses an apple vertically upward, releasing it 0.5 m above the floor. What is the maximum initial speed that can be given to the apple if it is not to touch the ceiling?
h = 3.75 – 0.5 = 3.25 m
h = v(o) •t - g•t^2/2
v = 0 = v(o) - g•t.
v(o) =sqrt(2•g•h) = sqrt (2•9.8•3.25) = 7.98 m/s
v(o) < 7.98 m/s
Well, if the apple touches the ceiling, we might have a real-life "fruit ceiling" situation! Anyway, to figure out the maximum initial speed, we can use a little physics. We know that the apple will reach its maximum height when its final vertical velocity is zero.
Using the equation vf^2 = vi^2 + 2ad (where vf is final velocity, vi is initial velocity, a is acceleration, and d is distance), we can calculate the maximum initial speed.
Given: vf = 0 m/s (since the apple reaches its maximum height), a = -9.8 m/s^2 (taking downward as the positive direction), and d = 3.75 m - 0.5 m = 3.25 m (the total distance the apple has to travel).
Plugging in the numbers, we have:
0^2 = vi^2 + 2(-9.8)(3.25)
Simplifying the equation:
0 = vi^2 - 63.55
Rearranging it:
vi^2 = 63.55
Taking the square root of both sides:
vi ≈ √63.55
So, the maximum initial speed that can be given to the apple (without touching the ceiling) is approximately equal to the square root of 63.55 m/s, or roughly 7.98 m/s.
Hope that didn't make you "a-peel-ing" over with laughter!
To find the maximum initial speed of the apple, we can use the principles of projectile motion.
Let's set up our coordinate system with the positive direction being upwards. We will take the initial position of the apple as the origin (0,0). The final height of the apple is the height of the ceiling, which is 3.75 m. The initial position of the apple is 0.5 m above the floor.
We need to find the initial speed at which the apple should be thrown upwards so that it does not touch the ceiling when it reaches its highest point.
We can use the equation for the maximum height in projectile motion:
H = (V_initial^2 * sin^2(theta))/(2 * g)
where:
H is the maximum height,
V_initial is the initial velocity of the apple,
theta is the angle of projection (which will be 90 degrees in this case since the apple is thrown vertically),
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
We know that the maximum height is 3.75 m, the angle of projection is 90 degrees, and the acceleration due to gravity is 9.8 m/s^2.
Plugging in these values into the equation, we can solve for V_initial:
3.75 = (V_initial^2 * sin^2(90))/(2 * 9.8)
Since sin(90) = 1, the equation simplifies to:
3.75 = (V_initial^2)/(19.6)
Multiplying both sides by 19.6, we get:
V_initial^2 = 3.75 * 19.6
V_initial^2 = 73.5
Taking the square root of both sides, we find:
V_initial = √73.5
V_initial is approximately 8.57 m/s.
Therefore, to prevent the apple from touching the ceiling, it should be thrown with a maximum initial speed of approximately 8.57 m/s.
To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the apple at any point in its motion is given by the sum of its kinetic energy (KE) and potential energy (PE).
At the beginning, when the apple is released 0.5 m above the floor, it has gravitational potential energy. As it is thrown vertically upward, it reaches its maximum height where its potential energy is at its maximum and its kinetic energy is minimum. At this point, all of the initial kinetic energy is converted into potential energy.
To determine the maximum initial speed, we need to find the potential energy at the maximum height and then equate it to the initial kinetic energy.
Let's solve it step by step:
Step 1: Determine the potential energy at maximum height.
The potential energy of an object near the Earth's surface is given by the formula:
PE = m * g * h
where m is the mass of the apple, g is the acceleration due to gravity, and h is the height. In this case, h is the distance between the maximum height and the floor, which is 3.75 m - 0.5 m = 3.25 m.
Step 2: Equate the potential energy at maximum height to the initial kinetic energy.
Since the apple is initially at rest when it is released, its initial kinetic energy is zero.
PE = KE_initial
m * g * h = (1/2) * m * v_initial^2
where v_initial is the initial velocity or speed of the apple. We can simplify the equation by canceling out mass on both sides:
g * h = (1/2) * v_initial^2
Now, substitute the known values:
9.8 m/s^2 * 3.25 m = (1/2) * v_initial^2
Step 3: Solve for v_initial.
Rearrange the equation to solve for v_initial:
v_initial^2 = 2 * g * h
v_initial^2 = 2 * 9.8 m/s^2 * 3.25 m
v_initial^2 = 63.7 m^2/s^2
v_initial = √(63.7 m^2/s^2)
v_initial ≈ 7.99 m/s
Therefore, the maximum initial speed that can be given to the apple is approximately 7.99 m/s.