Question

A hiker throws a stone from the upper edge of a vertical cliff. The stone's initial velocity is 25.0 m / s directed at 40.0° with the face of the cliff, as shown in the figure. The stone hits the ground 3.75 s after being thrown and feels no appreciable air resistance as it falls.

(a) What is the height of the cliff?

(b) How far from the foot of the cliff does the stone land?

(c) How fast is the stone moving just before it hits the ground?

Answer 1

a] Height H = ut + 0.5gt^2

= 25*cos 40 degree*3.75 + 0.5*9.8*3.75^2

= 140.7 m

b] d = u_horizontal *t = 25*sin 40 degree*3.75 = 60.26 m

c] Final V = - [25*cos 40 degree+9.8*3.75] j + 25*sin 40 degree i

= - 55.9 j + 16.07 i

speed = sqrt(55.9^2+16.07^2) = 58.16 m/s answer