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 vi = 18.2 m/s. The cliff is h = 49.5 m above a body of water as shown in the figure below.

(a) What are the coordinates of the initial position of the stone? (Do not assume that the student is the point of origin.)

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


(c) What is the appropriate analysis model for the vertical motion of the stone? (Select all that apply.)
5

constant velocity motion free fall motion g = 9.8 m/s2 no acceleration from gravity g = -9.8 m/s2

(d) What is the appropriate analysis model for the horizontal motion of the stone? (Select all that apply.)
6

constant velocity in the horizontal direction g = 9.8 m/s2 g = -9.8 m/s2 free fall motion no acceleration from gravity

(e) 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: vix, g, and t.)

(f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g and t.)

(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?

Explanation / Answer

given intial speed of the stone vi = 18.2m/s the cliff height = h = 49.5m a) at the intial postion the x,y cordinates of the stone is ( 0m, 49.5m) b) intial velocity of stone along horizontal direction si components vxi= 18.2m/s , vyi= 0m/s c) particle moving under constant accleration. d) particle under constant velocity e) vxf = vxi vyf = -gt becase vyi = 0 f) xf = vxi * t yf = yi- 1/2( g t2) g) time taken the stone to reach the water below is t = 2h/g      = 3.17s h) the velocity of projectile at any instnant is of time t is v = vxi2 + g2 t2     = 36m/s angel = - 60.10     = 36m/s angel = - 60.10

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