Question

A student stands at the edge of a cliff and throws a stone horlzontally over the edge with a speed of vi = 21.5 m/s. The cliff Is h = 50.3 m above a body of water as shown in the figure below. 

(a) What are the coordinates of the initial position of the stone? (Note that the origin is at the base of the cliff, where the x- and y-axes intersect.)

(b) What are the components of the initial velocity of the stone? 

(c) What is the approprlate analysis model for the vertical motion of the stone? 

 (d) What is the appropriate analysis model for the horizontal motion of the stone? 

(e) Write symbolic equations for thex and y components of the velocity of the stone as a function of time. (Use the following as necessary: v g, and t. Indicate the direction of the velocity with the sign of your answer.) 

(f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: v h, g, and t. Indicate the direction of the displacement with the sign of your answer.) 

(g) How long after being released does the stone strike the water below the cliff? 

(h) With what speed and angle of impact does the stone land?

Answer 1

a) the x component is since the initial point lies on the y axis at a height 'h' hence its coordinates are,

x_i = 0 m

y_i = 50.3 m

b) The y component of the velocity is zero, since it is thrown horizontally

v_ix = 21.5 m/s

v_iy =0 m/s

c) Particle under constant, non zero acceleration

only the force of gravity acts on the object vertically downwards

d) particle under constant speed,

since no net force is present in the horizontal direction

e) For the horizontal motion,(unaccelerated)

Since it is unaccelerated the velocity of the stone will not change with time

v_fx = v_ix = 21.5 m/s

For the vertical motion (accelerated)

initial velocity, v_iy = 0 m/s

using, v_fy = v_iy+ at

v_fy = gt

f)  For the horizontal motion,(unaccelerated)

x_f = v_fx*t

For the vertical motion (accelerated)

using (s= ut + 0.5at²)

y = 0*t + 0.5*g*t²

y = 0.5gt²

This is the displacement of the object as measured from the top of the cliff,

When measured from the origin

y_f = h- y

y_f = h - 0.5gt²

g) When the object hits the water, its displacement (y=h) or its displacement as measured from the origin is zero (y_f = 0)

0= h- 0.5gt²

t = sqrt(2h/g) = sqrt(2*50.3/9.81)

t = 3.202 s