Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Eric drops a 2.30 kg water balloon that falls a distance of 35.65 m off the top of abuilding. What is the kinetic energy at the bottom?

Sagot :

Givens.

• The mass of the ballon is 2.30 kg.

,

• The height is 35.65 m.

First, find the final velocity when the balloon is at the bottom. Use a formula that relates height, initial speed, final speed, and gravity.

[tex]v^2_f=v^2_0+2gh[/tex]

Where

• v_0 = 0 because the balloon starts from rest.

,

• Gravity is g = 9.8 m/s^2.

,

• h = 35.65 m.

Use all these magnitudes to find the final velocity v_f

[tex]\begin{gathered} v^2_f=0^2+2(9.8\cdot\frac{m}{s^2})(35.65m) \\ v^2_f=698.74m^2 \\ v_f=\sqrt[]{698.74m^2} \\ v_f\approx26.4(\frac{m}{s}) \end{gathered}[/tex]

Once we have the velocity at the bottom, find the kinetic energy using its formula.

[tex]K=\frac{1}{2}mv^2[/tex]

Where m = 2.30 kg, and v = 26.4m.s.

[tex]\begin{gathered} K=\frac{1}{2}\cdot(2.30\operatorname{kg})\cdot(26.4(\frac{m}{s}))^2 \\ K=1.15\cdot696.96J \\ K=801.5J \end{gathered}[/tex]

Therefore, the kinetic energy at the bottom is 801.5 J.