Question

Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The first setting compresses the spring by 1 and fires the ball a distance of 0.25 . The second setting compresses the spring by 2 and fires the ball a distance of 1 .Assuming that the relationship between spring compression and height is quadratic, how far will the ball go at the highest setting, which compresses the spring 3

Answer 1

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:

$\dfrac{1}{2}\times K \times x^2 = mgs +c$

By cross multiplying

$K \times x^2 = 2(mgs +c)$

$K \times x^2 = 2\times 9.81(ms) +2c$

$K \times x^2 = 19.62(ms) +2c$

$x^2 = A \times s+B$

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 :

$x^2 = A \times s+B$ 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