Answer:
a) 58.83 m b) 2262 N
Explanation:
The cord force constant K = 39 N/m, initial length of the cord L₀ = 33 m,
new length of the cord after the ball was just before the ball was pulled back L₁ = d + L₀ where d is the extension of the cord in meters
Weight of the ball = m g where m is the mass of the ball and g is acceleration due to gravity in m/s² = 75 × 9.8 = 735 N
loss of potential energy = equal to gain in elastic potential energy by the cord
loss in potential energy = mgh = 735 × ( d + L₀)
and the elastic potential gain by the cord = 1/2 × K ×d²
735(d + 33) = 0.5×39×d²
735(d + 33) = 19.5 d²
735 d + 24255 = 19.5 d²
rearrange the equation
-19.5 d² + 735 d + 24255 = 0
multiply both side of the equation with -1
19.5 d² - 735 d - 24255 = 0
solve the equation using quadratic formula
-b ± √(b² - 4ac) / 2a
where a = 19.5, b = -735, c = -24255
substitute this values into the formula
- (-735) ±√((-735²) - (4×19.5×-24255)) / (2×19.5)
735±√(540225 + 1891890) / (39)
735 ± √(2432115) / 39
735 ± 1559.5 / 39
(735+ 1559.5) / 39 or (735 - 1559.5) / 39
58.83 m or -21.1 m
the new length = 33 m+ 58.83 m = 91.83
or
the new length = 33 m + (-21) = 12
using the positive value
the ball's height above the water = 94 - 91.83 = 2.17 m
b) Tension in the cord = Kd using hooke's law
T = 39 × 58 = 2262 N
