(adsbygoogle = window.adsbygoogle || []).push({}); Projectile motion off a cliff at an angle (help plz)

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 y-axis 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

**Physics Forums - The Fusion of Science and Community**

# Projectile motion off a cliff at an angle (help )

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Projectile motion off a cliff at an angle (help )

Loading...

**Physics Forums - The Fusion of Science and Community**