PS2-2

# PS2-2 - Problem Set 2: Problem 2. Problem: A ball is...

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Unformatted text preview: Problem Set 2: Problem 2. Problem: A ball is launched perpendicular to a surface inclined to the horizontal at an angle φ as shown. It lands a distance measured along the incline. Determine the ball’s initial speed, V , as a function of , φ and gravitational acceleration, g . What is the numerical value of V if = 100 m and φ = 30o ? You can ignore effects of friction on the ball’s motion. ..... . . ..... . . .. .......... .. .. . . ............. . .. ... .. ................... ....................... . . . .... .. . . . . . . . .................... ........................... . ... . . . .. . . .. .. . ........................... . .. . . . . . .. . ................. .. . . ... . ........................... . . . .. .. .......................... . ... . ......... ..... .. ............................ . . . .......... .. .. .. .. ..... ... ........................... ... .. ...... . ............................ . .......... .. .. .. ........................................................................................................... .. . ... ... . ........................................... ........................................... .. . ........................... ........ . .. .. . . . ...... ........................... . . .. .. ....... ........ .. .. ... ........................... . . ................................. . ............ .. .. . ... . . ..... ................................ .. .. . ....... .................................. .. . ........................ .......... . . . .... ............................... .. .... . . .. .. ... .................. .. . . .... .. .... ............................ . ...... ...... . . .... .............................. . .... ............ . ............. ... . . . . .... .................................. .... ............................. . .. . . .. ... ..... . .... ................... ... ........................... .... . . ..... . . ... . .. ........................... ..................... .. . . .. . ............................ . ......... ..... .. . .. ... ...................... . .... .............................. . .... ........ . .......... .. . .. ....................... .. ....... .. .... .... .... ............. . .... ................................... .... .. .... ................................. ............. .. ... .... ............................ ...................... .... ........... . .... ........................... ..... . .. .... . .......................... .... . ..................... ..... . ..... ........................... . ................... .. ..... .. .......................... ... . . ........................... .... .................... . . . .......................... .................... ........................... ..................... .................. ... .. .. ................ ... z V φ x g = −g k • Solution: The first thing we must do is determine the x and z components of the initial velocity vector. For the geometry shown, clearly v = V i cos π π − φ + k sin −φ 2 2 = V [i sin φ + k cos φ] Therefore, the initial velocity components are x(0) = V sin φ ˙ and z (0) = V cos φ ˙ Because we ignore friction, the motion of the ball is subject only to the constant gravitational acceleration. Denoting time by t and letting t = 0 when the ball launches from the top of the incline, the equations and initial conditions governing the motion are d2 x = 0, dt2 d2 z = −g, dt2 x(0) = 0 and x(0) = V sin φ ˙ z (0) = 0 and z (0) = V cos φ ˙ Integrating twice and imposing the initial conditions, the solution to the equations for x and z is x(t) = V t sin φ When the ball lands at time τ , we know that x(τ ) = cos φ Therefore, we have V τ sin φ = cos φ =⇒ τ= cot φ V and z (τ ) = − sin φ and 1 z (t) = V t cos φ − gt2 2 30 Turning to the vertical position, we have CHAPTER 13. PARTICLE KINEMATICS 1 − sin φ = V τ cos φ − g τ 2 2 cot φ 1 cot φ =V cos φ − g V 2 V 2 g = cos φ cot φ − cot2 φ 2V 2 g = cot2 φ sin φ − 2V 2 Thus, there follows − sin φ tan2 φ = sin φ − g 2V 2 =⇒ 2 g = sin φ 1 + tan2 φ 2V 2 Using the trigonometric identity 1 + tan2 φ = sec2 φ = 1/ cos2 φ, a little rearrangement of terms yields V = cos φ For g 2 sin φ = 100 m and φ = 30o , the ball’s initial velocity is 9.807 m/sec 2 V = cos 30 o (100 m) = 27.1 2 sin 30o m sec ...
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## This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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