An athlete on a trampoline leaps straight up into the air with an initial speed of 9.3 m/s. Find the following:
(b) the speed of the athlete when she is halfway up to her maximum height
halfway up used up half the energy.
1/2 mv^2= 1/2 energy
so v= sqrt 1/2 of original speed.
To find the speed of the athlete when she is halfway up to the maximum height, we need to determine the height reached by the athlete first.
Let's denote the maximum height reached by the athlete as h_max. At the topmost point of the motion, the final velocity will be zero (v = 0 m/s). We can use the equation for motion in the vertical direction:
v^2 = u^2 - 2gh,
where:
v = final velocity (0 m/s at the top)
u = initial velocity (9.3 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = maximum height reached by the athlete
Rearranging the equation, we can solve for h:
h = (u^2 - v^2) / (2g).
Substituting the given values:
h = (9.3^2 - 0) / (2 * 9.8)
= 86.17 / 19.6
≈ 4.4 m.
Therefore, the maximum height reached by the athlete is approximately 4.4 m.
To find the speed of the athlete when she is halfway up, we need to find the height when she is halfway up on her upward trajectory. This height can be calculated as:
h_halfway = 0.5 * h_max
= 0.5 * 4.4
= 2.2 m.
Now, we can find the velocity at this height using the same equation:
v^2 = u^2 - 2gh,
where:
v = final velocity (to be determined)
u = initial velocity (9.3 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = height (2.2 m)
Rearranging the equation, we can solve for v:
v = √(u^2 - 2gh).
Substituting the given values:
v = √(9.3^2 - 2 * 9.8 * 2.2)
= √(86.49 - 43.12)
= √43.37
≈ 6.59 m/s.
Therefore, when the athlete is halfway up to the maximum height, her speed is approximately 6.59 m/s.
To find the speed of the athlete when she is halfway up to her maximum height, we need to use the principles of conservation of energy.
When the athlete is halfway up to her maximum height, she has reached the highest point of her motion and is momentarily at rest before starting to descend. At this point, all of the initial kinetic energy has been converted into potential energy.
Let's break down the problem step by step:
1. Determine the maximum height reached by the athlete. To do this, we need to use the formula for the maximum height reached by an object in free fall:
h = (v^2) / (2g)
where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth).
Plugging in the given values:
h = (9.3 m/s)^2 / (2 * 9.8 m/s^2)
Simplifying:
h = 44.89 m
2. Calculate the potential energy of the athlete halfway up to her maximum height. Potential energy is given by the formula:
PE = m * g * h
where PE is the potential energy, m is the mass of the athlete, g is the acceleration due to gravity, and h is the height.
Since we are only interested in the speed, which is independent of the mass, we can ignore the mass in this calculation. Therefore, the formula becomes:
PE = g * h
Plugging in the known values:
PE = 9.8 m/s^2 * (44.89 m / 2)
Simplifying:
PE = 219.26 J
3. Find the speed of the athlete when she is halfway up. At this point, all of the initial kinetic energy has been converted into potential energy, so the total energy remains constant throughout the motion. Therefore, the potential energy halfway up is equal to the initial kinetic energy:
KE = 1/2 * m * v^2
where KE is the kinetic energy, m is the mass of the athlete, and v is the speed.
Again, since we are only interested in the speed, we can ignore the mass in this calculation. Therefore, the formula becomes:
KE = 1/2 * v^2
Setting the initial kinetic energy equal to the potential energy halfway up:
1/2 * v^2 = 219.26 J
Solving for v:
v^2 = 219.26 J * 2
v^2 = 438.52 J
v = √(438.52 J)
Calculating the square root:
v ≈ 20.92 m/s
Therefore, the speed of the athlete when she is halfway up to her maximum height is approximately 20.92 m/s.