A student stands at the edge of a cliff and throws a stone horizontally over the

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of

The cliff is

above a body of water as shown in the figure below.

What are the coordinates of the initial position of the stone? What are the components of the initial velocity of the stone? What is the appropriate analysis model for the vertical motion of the stone? particle under constant speed particle under constant, nonzero acceleration What is the appropriate analysis model for the horizontal motion of the stone? particle under constant speed particle under constant, nonzero acceleration Write symbolic equations for the x and y components of the velocity of the stone as a function of time. (Use the following as necessary: v,x, g, and t. Indicate the direction of the velocity with the sign of your answer.) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g, and t. Indicate the direction of the displacement with the sign of your answer.) How long after being released does the stone strike the water below the cliff? s With what speed and angle of impact does the stone land?

Explanation / Answer

a) Initial position of stone according to the axis in the diagram ,

x = 0 m

y = h = 57.7 m

b) components initial velocity

vix = 14.5 m/s

viy = 0 m/s

c) particle under constant , non zero acceleration

as gravity is continuously acting on the particle.

d) particle under constant speed

as there is no acceleration in the x-direction

e) vfx = vix = 14.5

vfy = -g*t = -9.8*t

as acceleration due to gravity acts in negative y-direction

taking acceleration due to gravity = g = 9.8 m/s^2

f) xf = vfx*t = 14.5 t

yf = yi - 0.5*g*t^2 =57.7-(0.5*9.8*t^2 )= 57.7 -( 4.9 t^2 )

taking acceleration due to gravity = g = 9.8 m/s^2

g) for the stone to strike cliff ,

yf =0

57.7 -( 4.9 t^2 ) = 0

solving t = 3.4315 s

3.4315 s is the answer.

h) At t = 3.4315 s

vfx = 14.5

vfy = -9.8*t = -9.8*3.4315 = -33.6287 m/s

so magnitude of velocity = sqrt(vfx^2 + vfy^2) = sqrt( 14.5^2 + (-33.6287)^2) = 36.6215 m/s

v = 36.6215 m/s

theta = atan( vfy/vfx) = atan( -33.6287 / 14.5) = 293.324 degrees is the answer .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.