(1096) Problem 3: A student throws a water balloon with speed Vo from a height h
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(1096) Problem 3: A student throws a water balloon with speed Vo from a height h 1.92 m at an angle = 33° above the horizontal toward a target on the ground. The target is located a horizontal distance d 5.5 m from the student's feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position. 0 Otheexpertta.com 25% Part (a) what is the position vector. Rtarget that originates from the balloon's original position and terminates at the target? Put this in terms of h and d, and represent it as a vector using i and j Rargetdi-hj Correct! 25% Part (b) In terms of the variables in the problem. determine the time. t, after the launch it takes the balloon to reach the target. Your answer should not include t d ( vo cos() ) Correct! 25% Part (c) Create an expression for the balloon's vertical position as a function of time.)(), in terms oft. Vo, g, and y() = vo sin() t-( g2 ) t2 Correct! 25% Part d) Determine the magnitude of the balloon's initial velocity vo in meters per second by eliminating t fom the prevous two expressions Grade Summary Deductions Potential Vo-19.6 10% 90% S1 cos tan 78 9 HOME cotan asin0acos0 atan acotansinh0 coshtanh0cotanh0 Degrees O Radians Submissions Attempts remaining: 3 (5% per attempt) detailed view 1 2 3 0 END 5% BACKSPACE DEL CLEAR 5% Submit I give up! Hints: % deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback.Explanation / Answer
(d) Initial velocity in the horizontal direction (VX) = VoCos33
Initial velocity in the vertical direction (VY) = VoSin33
Now considering the horizontal motion
t = Distance./speed = d/VoCos33 = 5.5/VoCos33 = 6.56/Vo ----------(1)
Now considering the vertical motion
S = ut + (1/2)at2
-1.92 = (VoSin33)*(6.56/Vo) +(1/2)*(-9.81)*(6.56/Vo)2
-1.92 =3.571 -(211.08/Vo2)
(211.08/Vo2) = 3.571+1.92 = 5.491
Vo = 6.2 m/s
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