Answer:
The distance s of how far the ball will go at the highest setting = 2.25m
Explanation:
Let consider x to be the representative of the compression and the distance to be s
Recall that:
[tex]\dfrac{1}{2}\times K \times x^2 = mgs +c[/tex]
By cross multiplying
[tex]K \times x^2 = 2(mgs +c)[/tex]
[tex]K \times x^2 = 2\times 9.81(ms) +2c[/tex]
[tex]K \times x^2 = 19.62(ms) +2c[/tex]
[tex]x^2 = A \times s+B[/tex]
Thus, for the first setting
x = 1 , s = 0.25
for the second setting
x = 2, s = 1
1 = 0.25A + B --- (1)
4 = A + B ----- (2)
From (1); let B = 1 - 0.25A and substitute it into (2)
4 = A + 1 - 0.25 A
4 - 1 = A - 0.25 A
3 = 0.75 A
A = 3/0.75
A = 4
From (2)
4 = A + B
4 = 4 + B
B = 4 - 4
B = 0
Therefore, for the highest setting, where x = 3
Then :
[tex]x^2 = A \times s+B[/tex] will be:
3² = 4s + 0
9 = 4s
s = 9/4
s = 2.25 m
∴
The distance s of how far the ball will go at the highest setting = 2.25m
