Projectile motion off a cliff at an angle (help )

[Wraith09]

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

Hi all,

I am new to the forum, 1st year chemical engineering student. I am struggling with this problem:

Homework Statement

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

Homework Equations

s_y = s_yo + v_yot + 0.5a_yt²

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 - 0.5at²

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_oyt + 0.5(a)t²

s_y = s_oy +(x / (cos 37')t) (sin37') t + 0.5(a)t²

57.516m = 40 + 75 tan37' +0.5(-9.8m.s^-2)t²

t² = 8.16
t = 2.85s

That is the answer I get... which is wrong as its not listed in the MC's

 

[rl.bhat]

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

 

[Wraith09]

Thank you, that's the one thing I missed!