qz1sol_3706Hs11 - θ determines the flight time let t max...

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TA: Tomoyuki Nakayama Tuesday, January 25, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 1 (Homework Set #2) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A ball is to be shot from level ground with a certain speed. The figure below shows the range R it will have versus the launch angle θ 0 . a) What is the launch speed? According to the graph, the maximum range is 300 m. The maximum range is achieved when the projection angle is 45º. The equation for horizontal range yields R max = v 0 2 /g v 0 = (R max g) = 54.2 m/s b) The value of
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Unformatted text preview: θ determines the flight time; let t max represent the maximum flight time. What is the least speed the ball will have during its flight if θ is chosen such that the flight time is 0.750 t max ? When the ball comes back to the ground level, the vertical displacement is zero. The flight time is 0 = v sin θ t-(1/2)gt 2 t f = 2v sin θ /g The sine function takes its maximum value (= 1) when θ is 90º. Thus the maximum flight time is t max = 2v /g Since we have to choose the projection angle so that t f = 0.75t max , the angle is 2v sin θ /g = 0.75(2v /g) sin θ = 0.75 θ = 48.6º The speed of the ball takes the minimum value at the peak of the projectile motion, where y component of the velocity is zero. Therefore, the least speed is v min = v 0x = v cos θ = 35.8 m/s...
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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