Hello there, I am trouble figuring out how to solve this even though thereis a s

Question

Hello there, I am trouble figuring out how to solve this even though thereis a solution on cramster I do not understand the answer in part a)especially the equation iii ? Can someone please explain thosesteps? Thank you so much!!!! Here is the question:
A 60 kg soccer player jumpsvertically upwards and heads the 0.45 kg ball as it is descendingvertically with a speed of 25 m/s. If the player was moving upward with aspeed of 4.0 m/s just before impact, what will be the speed of theball immediately after the collision if the ball reboundsvertically upwards and the collision is elastic?
Here is howcramster expert has solved it:
0.5 m(uball)^2 + 0.5 Mu^2=0.5 m(v ball)^2+0.5 MV^2 ....equationi and m (uball)+MU=m(Vball)+MV...equation ii
fromequation i and ii u+U=Uball+Vball 4+V=-25+Vball V=Vball=2b.... equation iii
60V +0.45Vball=226.75....equation iv
fromequation iii and iv Vball=1.799m/s (up)
Hello there, I am trouble figuring out how to solve this even though thereis a solution on cramster I do not understand the answer in part a)especially the equation iii ? Can someone please explain thosesteps? Thank you so much!!!! Here is the question:
A 60 kg soccer player jumpsvertically upwards and heads the 0.45 kg ball as it is descendingvertically with a speed of 25 m/s. If the player was moving upward with aspeed of 4.0 m/s just before impact, what will be the speed of theball immediately after the collision if the ball reboundsvertically upwards and the collision is elastic?
Here is howcramster expert has solved it:
0.5 m(uball)^2 + 0.5 Mu^2=0.5 m(v ball)^2+0.5 MV^2 ....equationi and m (uball)+MU=m(Vball)+MV...equation ii
fromequation i and ii u+U=Uball+Vball 4+V=-25+Vball V=Vball=2b.... equation iii
60V +0.45Vball=226.75....equation iv
fromequation iii and iv Vball=1.799m/s (up) Hello there, I am trouble figuring out how to solve this even though thereis a solution on cramster I do not understand the answer in part a)especially the equation iii ? Can someone please explain thosesteps? Thank you so much!!!! Here is the question:
A 60 kg soccer player jumpsvertically upwards and heads the 0.45 kg ball as it is descendingvertically with a speed of 25 m/s. If the player was moving upward with aspeed of 4.0 m/s just before impact, what will be the speed of theball immediately after the collision if the ball reboundsvertically upwards and the collision is elastic?
Here is howcramster expert has solved it:
0.5 m(uball)^2 + 0.5 Mu^2=0.5 m(v ball)^2+0.5 MV^2 ....equationi and m (uball)+MU=m(Vball)+MV...equation ii
fromequation i and ii u+U=Uball+Vball 4+V=-25+Vball V=Vball=2b.... equation iii
60V +0.45Vball=226.75....equation iv
fromequation iii and iv Vball=1.799m/s (up)

Explanation / Answer

A 60 kg soccer player jumpsvertically upwards and heads the 0.45 kg ball as it is descendingvertically with a speed of 25 m/s. If the player was moving upward with aspeed of 4.0 m/s just before impact, what will be the speed of theball immediately after the collision if the ball reboundsvertically upwards and the collision is elastic? The mass of the player M = 60 kg The mass of the ball m = 0.45 kg The speed of the player u = 4.0 m/s The speed of the ball v = - 25 m/s The speed of the player after impact the ball u' = 0 The speed of the after impact = v' = ? According to conservation of momentum we have      M u + m v = M u' + m v'      60 * 4 - 0.45 * 25 = 0 + 0.45v'      v ' = 508.33 m/s

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