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 = 53.5 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.) xi = m yi = m (b) What are the components of the initial velocity of the stone? vix = m/s viy = m/s (c) What is the appropriate analysis model for the vertical motion of the stone? (Select all that apply.) constant velocity motion g = 9.8 m/s2 free fall motion g = ?9.8 m/s2 no acceleration from gravity (d) What is the appropriate analysis model for the horizontal motion of the stone? (Select all that apply.) no acceleration from gravity constant velocity in the horizontal direction free fall motion g = ?9.8 m/s2 g = 9.8 m/s2 (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.) vfx = vfy = (f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g and t.) xf = yf = (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? speed m/s direction

Explanation / Answer

h(t) = -(9.81/2) t^2 + 35.2 m i t^2 = 35.2 / 4.905 t = 2.6788 seconds the horizontal departure velocity has nothing to do with the vertical descent 3 years ago Report Abuse Answerer 2 y = yo + vo*t - (1/2)gt^2 initial vertical velocity is vo=0 initial height yo =0 final height y=-35.2 gravitational acceleration g=9.8 -35.2=0+0-(1/2)(9.8)(t^2) t=2.68seconds 3 years ago Report Abuse Answerer 3 y = yo + vo*t - (1/2)gt^2 = 35.2 + 16t - 4.9t^2 with one positive root at t = 4.771 s "The cliff is 35.2 m above a flat horizontal beach" This is the vertical distance. The speed 16 m/s is the magnitude of the stone's speed, it can be taken apart in horizontal and vertical direction (rcos(t), rsin(t)) if it needs to. The second answer included g but is woefully unaware that the horizontal distance is not dependent on the acceleration due to gravity Edit: I can see why the second answer would be valid if the question meant to say that the stone was thrown at a 0 degree angle. 16sin(0) = 0 and 16cos(0) = 16. In which case h(t) = -gt^2 /2 + ho would be the answer

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