Question

Help! will give medals if you can explain:

A soccer ball m=0.65kg is dropped from height h=4.9m bounces off the grass. The ball loses E=3.4J of energy to the environment during the bounce.

Write an expression for high (final height) the soccer ball bounces.

Answer 1

@TheSmartOne @IrishBoy123

Answer 2

@.Sam. @Elsa213

Answer 3

Initial Energy (Potential) : \(U_1 = mgh_1\)

When it hits the grass, Energy (Kinetic) is : \(T_1 = \frac{1}{2} m v_1^2 = U_1\)

Energy immediately after collision (Kinetic): \(T_2 = T_1 - 3.4 = U_1 - 3.4\)

Once it rises and converts all that kinetic into potential energy, we say that:

\(U_2 = T_2 = U_1 - 3.4\)


\(\implies mgh_1 = mgh_o - 3.4\)

\(\implies h_1 = h_o - \dfrac{3.4}{mg}\)

\(h_1 = h_o - \dfrac{3.4}{0.65 * 9.81}\) m

Answer 4

@IrishBoy123 , what is the initial expression you used for h final?
h Final= h initial-(E/(mg))?

Answer 5

\[ho-(\frac{ E }{ mg })m\] keeps coming up as incorrect?

Answer 6

Yes

\(h_f = h_i - \dfrac{\mathbf{Energy loss}}{0.65 * 9.81}\) m

Initial Energy (which is Potential) : \(E_i = U_i = mgh_i\)

When it hits the grass, Energy (which is Kinetic) is : \(E_i = T_i = \frac{1}{2} m v_i^2 = U_i\)

Energy immediately after collision (Kinetic): \(E_f = T_f = T_i - 3.4 = U_i - 3.4\)

Once it rises and converts all that kinetic into potential energy, we say that:

\(E_f = U_f = T_f = U_i - 3.4\)


\(\implies mgh_f = mgh_i - 3.4\)

\(\implies h_f = h_i - \dfrac{3.4}{mg}\)

\(h_f = h_i - \dfrac{\mathbf{Energy loss}}{0.65 * 9.81}\) m

Answer 7

oh, :-(

the m is metres, so sorry

Answer 8

\(h_f = h_i - \dfrac{\mathbf{Energy loss}}{0.65 * 9.81}\)


putting the unit in....
\(h_f = \left( h_i - \dfrac{\mathbf{Energy loss}}{0.65 * 9.81}\right) \mathbf{m}\)

mea culpa :(

Answer 9

i messed up with my paeantheses, thank you!!! i had it right the whole time, i just never used the correct placement of parantheses, thnak you!

Answer 10

if i could give you a 2nd medal, I would. thank you :)

Answer 11

lol!!

:)