from what height must a heavy elastic ball be dropped on a floor ,so thatafter reboundingthrice it will reach a height of 16 m?take e=(0.5)
After the 1st rebounce the speed is v/2, after the 2nd rebounce – v/4,
after the third rebounce –v/8, =>
m(v/8)²/2=mgh₁
v=sqrt(128gh₁) =141.7 m/s
mv²/2=mgh
h= v²/2g=141.7²/2•9.8=1024 m
Well, let's see if I can help you with your elastic ball dilemma. If the ball rebounds thrice, it means it's going to bounce back twice after being dropped. So, we need to consider the height it reaches after each rebound.
Let's assume the initial height the ball is dropped from is "H." After the first bounce, it reaches a height of 0.5H (according to the given e = 0.5). After the second bounce, it reaches a height of (0.5H * 0.5) = 0.25H. Finally, after the third bounce, it reaches a height of (0.5H * 0.5 * 0.5) = 0.125H.
Now, we know that the ball needs to reach a height of 16m after the third bounce. So, we can set up an equation:
0.125H = 16
Now, we just need to solve for H:
H = 16 / 0.125
H = 128
Therefore, the heavy elastic ball must be dropped from a height of 128 meters for it to reach a height of 16 meters after rebounding thrice.
Now, don't go dropping anything else from that height, unless it's a joke book!
To solve this problem, we need to use the concept of elastic potential energy. The mechanical energy of a ball is conserved during an elastic collision, where the kinetic energy is transferred to potential energy.
Given:
- Desired height after three rebounds: h = 16 m
- Coefficient of restitution: e = 0.5
We can use the following equation to determine the initial height from which the ball should be dropped:
h = (e^n) * h0
Where:
- h0 is the initial height
- n is the number of rebounds
In this case, the ball rebounds three times (n = 3):
16 = (0.5^3) * h0
To simplify the equation, let's calculate (0.5^3):
0.5^3 = 0.5 * 0.5 * 0.5 = 0.125
Now, substitute this value back into the equation:
16 = 0.125 * h0
To isolate h0, divide both sides of the equation by 0.125:
h0 = 16 / 0.125
h0 = 128 m
Therefore, the heavy elastic ball must be dropped from a height of 128 meters in order to reach a height of 16 meters after rebounding three times.
To find the height from which the heavy elastic ball must be dropped, we can use the concept of elastic potential energy.
The elastic potential energy of the ball is given by the equation:
EPE = (1/2)ke^2
where EPE is the elastic potential energy, k is the spring constant, and e is the elongation or compression of the spring.
In this case, we need to consider the compression of the ball due to its rebounding. Since the ball rebounds thrice, the total compression (e) will be equal to three times the original compression (e).
The equation for the total compression is given by:
te = 3e
where te is the total compression.
Since we know the height (h) at the final rebound is 16 m, we can equate the elastic potential energy at the initial height (h0) to the elastic potential energy at the final height (hf).
Since the ball is dropped, the initial potential energy (PE) is equal to mgh0, where m is the mass, g is the acceleration due to gravity, and h0 is the initial height.
Therefore, the initial elastic potential energy (EPE0) is given by:
EPE0 = mgh0
The final elastic potential energy (EPEf) is given by:
EPEf = mghf
Since both EPE0 and EPEf are equal, we can equate them:
mgh0 = mghf
Cancelling out the mass and acceleration due to gravity, we get:
h0 = hf
Since we need to find the initial height (h0), we can substitute h0 with hf:
h0 = 16 m
Therefore, the heavy elastic ball must be dropped from a height of 16 meters in order to reach a height of 16 meters after rebounding thrice.