Dynamics_Part13

Dynamics_Part13 - Problem Set 2: Problem 3. Problem: A...

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Unformatted text preview: Problem Set 2: Problem 3. Problem: A projectile is launched with initial speed nV as seen by a stationary observer at an angle to the horizontal from a height h above the ground. The object is to have the projectile strike the center of a vehicle as shown. The center of the vehicle (whose top is also at height h) is at a horizontal distance L from the launch point at time t = 0 when the projectile is launched. The vehicle's speed at that time is vx (0) = V . The vehicle accelerates horizontally with a = 1 g, where g is gravitational 2 acceleration. Ignore effects of friction on the ball's motion. Verify that n satisfies a quadratic equation involving the dimensionless quantity gL/V 2 . HINT: Set the vertical origin of your coordinate system at a distance h above the ground. . . . . . . . . . . . . . . . . . . . . . ... ... . ... . .. .. . . ......... . . . .. . . . . . . . . . . . .. . . . ... . ... .... . . .. .. ..... .. . . . .. ... . .. ... . . .. .. .... . .... . .. . .. ... . . ... . . .. .. .. .. .... . .. ..... .. ..... . .. ... . ... . .. . 1 . ..................... . ......... ..... . ................... .. ................ ... . ... ......... .. . . . .......................................................... . ......................................................... .... . ........... ...... ............ ................ . ............................................................................. ................ .. .. ........... ............ ............ 2 .. .............................................................. x .. . . . . . . ............................................................................... . . ................. ................................................................ ........... .. .. ............. .............. .......... .. .. .......... ............................................................. ......... . . . . . .. . .. .. ............................................................. . .. . ........... .......... ............... .... ... . ............ ..... . . .......................................................... . . .. . .. .. . . .. . .. .. .. .......................................... . ........... ... ........................................... ............ .............................................. ............ ...... .............. . ............................................................................................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................................................................................. ...... . ............ ... . . ... ................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................................................ g nV a= g v (t) v v Solution: To solve this problem, we first solve for the projectile's position as observed by a stationary observer. Then, we compute the accelerating vehicle's position, again as observed by a stationary observer. Equating the projectile and vehicle positions yields the desired equation. Denote the projectile's coordinates by xB and zB . Then, the equations of motion and initial conditions are d2 xB = 0; dt2 d2 zB = -g; dt2 The solution is xB (t) = nV t cos and xB (0) = 0, xB (0) = nV cos zB (0) = 0, zB (0) = nV sin 1 zB (t) = nV t sin - gt2 2 The projectile strikes the vehicle when zB = 0, i.e., when nV t sin = 1 2 gt 2 = t= 2nV sin g Thus, when the projectile strikes the vehicle, its horizontal coordinate, xB , is xB = 2n2 V 2 sin cos g Denoting the vehicle's position by xA , its motion is governed by 1 d2 xA = g; dt2 2 The solution is xA (0) = L, xA (0) = V 1 xA (t) = L + V t + gt2 4 ...
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