This gives 0 1 2 gt 2 h v y t h with solutions t h 0

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This gives 0 = - 1 2 gt 2 h + v 0 y t h , with solutions t h = 0 (not relevant, this is when ball is thrown) or t h = 2 v 0 y /g . Doubling the velocity ~v 0 doubles the y component v 0 y which doubles the time t h = 2 v 0 y /g . So the correct answer is twice 4 sec = 8 sec, answer (a). Note that the angle and the horizontal motion here play no role. A more explicit way to see this is to write t B = 2 v 0 y,B /g , t A = 2 v 0 y,A /g , so the ratio is t A /t B = v 0 y,A /v 0 y,B = ( v 0 ,A sin(30 )) / ( v 0 ,B sin(30 )) v 0 ,A /v 0 ,B . 5. A bicycle wheel is spinning so fast it completes exactly four turns each second. The breaks are applied softly, so that the wheel spins down with constant deceleration, coming to rest in 220 s. How many turns did the wheel complete from the instant the breaks were first applied until it came to rest? (a) 440 4 (b) 220 (c) 2800 (d) 880 (e) 1800 π 4
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ANS: Circular motion kinematics with constant acceleration is just like one dimensional kinematics with constant acceleration: θ ( t ) = 1 2 αt 2 + ω 0 t + θ 0 and ω ( t ) = αt + ω 0 . At t = 0 the angular velocity is ω 0 = four turns in one second = 4(2 π ) rad/s, and slows down with constant acceleration so that when t = t s = 220 s the wheel stops, that is ω ( t s ) = 0. This gives 0 = αt s + ω 0 or α = - ω 0 /t s . The angular displacement after this time is Δ θθ ( t s ) - θ 0 = 1 2 αt 2 s + ω 0 t s = 1 2 ( - ω 0 /t s ) t 2 s + ω 0 t s = 1 2 ω 0 t s . Plugging in we have Δ θ = 1 2 (8 π )(220) rad. Divide by 2 π to get number of revolutions, Δ θ/ 2 π = 1 2 (8 π )(220) / 2( π ) = 440. 6. You are standing on the south shore of a river that flows from East to West at 5 m/s. You can swim in standing water at a speed of 8 m/s. In what direction should you swim to get across the river directly across your starting point? (a) 39 East of North 4 (b) 51 East of North (c) 32 East of North (d) 58 East of North (e) Directly north ANS: Addition of velocities. Let ~v WG = velocity of Water (river) relative to Ground (the shore), ~v PW = velocity of person (“you”) relative to Water and ~v PG = velocity of person relative to Ground. Then ~v PG = ~v PW + ~v WG . Graphically ~v PW ~v WG ~v PG This forms a triangle similar to: 8 sin θ = 5 8 5 | ~v PG | θ So θ = sin - 1 (5 / 8) = 39 .
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