A ball with a mass of 140 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 21 (kg)/s^2 and was compressed by 3/8 m when the ball was released. How high will the ball go?

Oct 21, 2017

1.076 m

Explanation:

Potential stored in spring = (kx^2)/2 = (21 kg.s^-2 × (3/8 m)^2)/2 = 189/128J

Kinetic energy gained by ball $= \frac{m {v}^{2}}{2}$
= (140 cancel(g) × 10^-3 (kg)/cancel(g) × v^2)/2 = (0.07 kg) × v^2

According to law of conservation of energy
Potential energy in spring = Kinetic energy gained by ball

189/128 J = (0.07 kg) × v^2

${v}^{2} = 21.09375 \frac{J}{k g}$

H = v^2/(2g) = (21.09375 J/(kg)) / (2 × 9.8 ms^-2)= 1.076 m