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Projectile motion off a cliff at an angle (help ) 
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#1
Feb1110, 02:07 AM

P: 3

Hi all,
I am new to the forum, 1st year chemical engineering student. I am struggling with this problem: 1. The problem statement, all variables and given/known data A stone is thrown at an angle of 37' from the top of a cliff which is 40m above the surface of a lake. The stone hits the lake's surface at a point which is 75m horizontally from the base of the cliff. Find the total time of flight of the stone: MCQ: a) 4.44s b) 4.00s c) 3.80s d) 3.60s e) 3.40s 2. Relevant equations s_{y} = s_{yo} + v_{yo}t + 0.5a_{y}t^{2} 3. The attempt at a solution I drew a free body diagram for the statement incl. a triangle with an angle of 37'. The hypotenuse of the triangle is speed (v_{o}) and its two sides are v_{ox} and v_{oy}. I also calculated S_{y} using Cosө = adj / hyp and then sinө = opp / hyp to get 56.516m To find v_{oy}: sin 37' = v_{oy} / v_{o} v_{oy} = v_{o} sin 37' To find v_{ox}: cos 37' = v_{ox} / v_{o} v_{ox} = v_{o} cos 37' I relate v_{ox} with dislpacement x over a certain period through the eq: v_{ox} = x / t x = (v_{ox}) t I relate motion along the yaxis to the displacement in this direction over a certain period via the eq: y = (v_{0y})t  0.5at^{2} Sustituting eqs: v_{o} = x / t v_{o} cos 37' = x / t v_{o} = x / (cos 37') t Substituting eqs: s_{y} = s_{oy} + v_{oy}t + 0.5(a)t^{2} s_{y} = s_{oy} +(x / (cos 37')t) (sin37') t + 0.5(a)t^{2} 57.516m = 40 + 75 tan37' +0.5(9.8m.s^{2})t^{2} t^{2} = 8.16 t = 2.85s That is the answer I get... which is wrong as its not listed in the MC's 


#2
Feb1110, 03:13 AM

HW Helper
P: 4,435

Hi Wraith09, welcome to PF.
Measure the displacement from the starting point. So so(y) = 0 and s(y) = 40 m. 


#3
Feb1110, 03:27 AM

P: 3

Thank you, thats the one thing I missed!



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