Arnoldo Budzynski: There is no force in the horizontal direction, so the horizontal coordinate is ignorable--we only need to worry about vertical velocity which is given by:vy = v sin thetaCONSERVATION OF ENERGY:initial KE (vertical) = 1/2 m (v sin theta)^2= final PE = mghDo the algebra to solve for height:h = (v sin theta)^2 / 2g = about 61m...Show more
Agustina Stimmel: At the ground level it has only kinetic energy = .5 m v^2 When it reached the maximum height it has horizontal velocity of V cos θ. Therefore it has a potential energy of mgh and a kinetic energy of 0.5 m [V cos θ] ^2Total energy = mg H + 0.5 m [V cos θ] ^2By conservation of energy:0.5 m V^2 = mg H + 0.5 m [V cos θ] ^2Canceling m through out 0.5 V^2 = g H + 0.5 [V cos θ] ^2 Multiplying by 2 V^2 = 2g H + [V cos θ] ^22g H = V^2 {1â"€ [cos θ] ^2} = V^2* [sin θ] ^2H = V ^2* [sin θ] ^2 / 2g. H = 61.22m...Show more
Lawana Neemann: It become going 2 m/s downward. we are going to call that! -2. Then it bounced and started going a million.6 m/s upward. we are going to call that +a million.6. the adaptation between taking place at 2 m/s and going up at a million.6 m/s is an entire of three.6 m/s. The ball weighs a million kg, so as that's 3.6 kgm/s. If the ball weighed 2 kg, the substitute in momentum would be 7.2 kgm/s. the comparable is going for if the a million kg ball become taking place at 4 m/s and then bounced up at 3.2 m/s....Show more
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