A basketball player crouches while waiting to jump for the ball. He lowers his center of gravity is 0.30 m. When he jumps for the ball, his center of gravity reaches a height of 0.90 m above its normal position.(a) Calculate the velocity of the player when his feet left the ground. (b) Calculate the acceleration he produced to achieve this velocity. (c) What force did he exert on the floor if his mass is 110 kg?
(a)
(1/2) m v^2 = m g h = m (9.8)(.9)
v^2 =2*9.8*.9
v = 4.2 m/s
(c)
work done to get from -.3 m to 0 meters where feet leave ground = F*.3 = change in potential + change in kinetic
F * .3 = m g (.3) + .5 m (4.2)^2
.3 F = 110 (.3*9.8 + .5*17.64)
F = 4312 N
(b)
F - mg = m a
4312 - 1078 = 110 * a
a = 29.4 m/s^2
wow, 3 g !
alternate (b)
we have v = 4.2
so average v during acceleration = 2.1
so time to go up .3 m = .3/2.1 = .143
change in v/time = a = 4.2/.143 = 29.4 again
Thank you Damon, It's appreciated
(a) Well, let's do some calculations and see how high we can jump with these numbers. If the basketball player's center of gravity is lowered by 0.30 m and reaches a height of 0.90 m above its normal position, that means he jumped a total of 0.90 + 0.3 = 1.20 m.
Now, we can use the equation of motion s = ut + 0.5at^2, where s is the height jumped, u is the initial velocity, a is the acceleration, and t is the time taken. We can assume that the time taken to reach the top is the same as the time taken to come back down (assuming no air resistance).
Using this equation and substituting the values, we get 1.20 = 0 + 0.5a(t/2)^2. Simplifying, we get 1.20 = 0.125at^2.
Now, let's tackle the acceleration. We'll need to use another equation, v = u + at, where v is the final velocity. Since his initial velocity is 0 when he jumps, the equation becomes v = at. From here, we can substitute this equation into the previous one to eliminate t.
1.20 = 0.125a(at)^2. Simplifying further, we get 1.20 = 0.125a^2 t^2.
Now, let's solve for t. Taking a square root of both sides, we get t = √(1.20 / 0.125a^2).
Now, let's substitute this value back into the equation v = at to find the initial velocity. v = a √(1.20 / 0.125a^2)
Simplifying this equation, we get v = √((8.0 / 0.125) a). Don't worry, this looks complicated, but we're almost there.
To find the velocity when his feet left the ground, we can assume the time taken to be the same as the time taken to reach the top. So, the velocity at takeoff v takeoff = a √(1.20 / 0.125a^2).
But wait, there's a conveniently cancelled-out 'a' in the numerator and denominator. So, v takeoff = √(1.20 / 0.125) m/s.
(b) Now, let's move on to the acceleration. We can use the equation v = u + at again, but this time, we already know the initial velocity and the final velocity is 0 at the top. So, the equation becomes 0 = u + at.
Simplifying, we get -u = at.
We can substitute the value of 'u' from the previous equation (v takeoff = √(1.20 / 0.125)) to eliminate 'u'. So, -√(1.20 / 0.125) = at.
Simplifying further, we get a = -√(1.20 / 0.125) / t.
Now, let's substitute the value of 't' that we found earlier (t = √(1.20 / 0.125a^2)). Simplifying the equation, we get a = -√(1.20 / 0.125) / √(1.20 / 0.125a^2).
As you can see, 'a' conveniently cancels out again, leaving us with a = -√(1.20 / 0.125) / √1.20.
(c) Finally, let's calculate the force exerted on the floor. We can use Newton's second law of motion, F = ma. Therefore, F = 110 kg * a.
Substituting the value of 'a' that we found earlier, we get F = 110 kg * (-√(1.20 / 0.125) / √1.20).
And voila, we have the force exerted on the floor. Remember, negative signs don't always mean bad things; they just indicate the direction of the force.
To solve this problem, we can use the principles of kinematics and dynamics. Here's how we can find the answers to each part of the question:
(a) Calculate the velocity of the player when his feet left the ground:
We can use the principle of conservation of energy to find the velocity of the player when his feet left the ground. At the start, the player's gravitational potential energy is converted into kinetic energy as he jumps. We can use the equation:
Kinetic Energy (K.E.) = Gravitational Potential Energy (G.P.E.)
The kinetic energy is given by the formula:
K.E. = (1/2) * mass * velocity^2
The gravitational potential energy is given by the formula:
G.P.E. = mass * gravity * height
Assuming the player's initial position is the reference point for potential energy, the change in height is:
Δh = height - initial height
= 0.90 m - 0.30 m
= 0.60 m
Setting the two energies equal, we have:
(1/2) * mass * velocity^2 = mass * gravity * Δh
Cancel the mass from both sides:
(1/2) * velocity^2 = gravity * Δh
Now solve for the velocity:
velocity^2 = (2 * gravity * Δh)
velocity = √(2 * gravity * Δh)
Substituting the known values:
velocity = √(2 * 9.8 m/s^2 * 0.60 m)
velocity ≈ 3.92 m/s
Therefore, the velocity of the player when his feet left the ground is approximately 3.92 m/s.
(b) Calculate the acceleration he produced to achieve this velocity:
To find the acceleration, we can use the formula:
velocity = initial velocity + acceleration * time
In this case, the initial velocity is 0 (since the player starts from rest), and the time it takes to achieve this velocity is not provided. However, we can assume the time it takes to achieve this velocity is very short (instantaneous), so the player's acceleration can be calculated as:
acceleration = velocity / time
Since the time is assumed to be very small, we can take it as zero. Therefore, the acceleration can be considered as infinite or undefined because it is a non-continuous, instantaneous change.
(c) What force did he exert on the floor if his mass is 110 kg:
To find the force exerted by the player on the floor, we can use Newton's second law of motion:
Force = mass * acceleration
Since we have already determined that the player's acceleration is undefined or infinite, we cannot calculate the force using conventional means. However, it is important to note that during the jump, the player exerts a force on the floor greater than their weight (mg) in order to propel themselves upwards. This force is caused by the large upward impulse delivered to the floor from the player's legs and is greater than the player's weight to overcome gravity.
It is also important to consider that during the player's jump, their force on the floor is not constant. It varies over time, decreasing after the jump starts until the feet leave the ground. Therefore, obtaining a specific value for the force exerted by the player on the floor would require knowing the time and specifics of the player's force-time profile during the jump.