A hiker throws a stone from the upper edge of a vertical cliff (see the figure b

Question

A hiker throws a stone from the upper edge of a vertical cliff (see the figure below). The stone's initial velocity is 25.0 m/s, directed at an angle of 40.0° with respect to the face of the cliff as shown in the figure. The stone hits the ground 3.75 s after being thrown and experiences no appreciable air resistance as it falls. (a) What is the height of the cliff? (b) How far from the base of the cliff does the stone land? (c) What is the stones speed right before it hits the ground? (d) How far from the foot of the cliff will the stone land if the mass of the stone was 100g?

Explanation / Answer

Vo =25 m/s, t =3.75 S, theta = 40 degrees

(a) from kinematic equations

s = ut+(1/2)at^2

y = ((25)cos(40)*3.75)+(0.5*9.8*3.75*3.75)

y =141 m

(b) x = vox*t

x = 25*sin(40)*3.75 = 60.26 m

(c) vx = 25*sin(40) = 16.07 m/s

from kinematic equation

vy = voy+at

vy = 25*cos(40) +9.8*3.75 =55.9 m/s

v = (16.07^2+55.9^2)^0.5

v = 58.16 m/s

(d) it does not depend on mass

x = 60.26 m

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