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

Question

A student stands at the edge of cliff and throws a stone horizontally over the edge with speed of 18.0 m/s. The cliff is 50.m above a flat, horizontal beach. a. what are the coordinates of the initial position of the stone? b. what are the coordinates of the initial velocity? c. Write the equations for the x and the y components of the velocity of the stone with time. D. write the equations for the position of the stone with time using the coordinates? Acceleration is equal to g. e. how long after being released does the stone strike the beach below the cliff? f. with what speed and angle of impacty does the stone land?

Explanation / Answer

find the vertical velocity, and the time needed for a vertical free fall from 54 m:

for the time use
s = 1/2*gt^2
t = ?(2s/g) = ?(2*50/9.8) = 3.19 s <-- time in which stone strikes beach

and the vertical velocity is then
v = gt = 9.8*3.19= 31.26 m/s

the total speed is the vector sum of vertical and horizontal velocities (since they are perpendicular, you can use the Pythagoras theorem):
v tot = ?(31.26^2 + 18^2) = 36.07 m/s <-- total velocity

the angle iat which the stone hits the ground is the angle of the sum vector of the two velocities:
angle = tan^-1(31.26/18) = 60.06

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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