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 degrees 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?

It can be defined as the motion of any object or body when thrown from the earth's surface and follows any curved path under the effect of the gravitational force of the earth.

The horizontal component of the velocity during any projectile motion is constant and is given by vcosθ.

As given in the problem 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 degrees 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

The horizontal component of the velocity is given by vsinθ

Vx = vsinθ

= 25×sin40°

=16.06 m/s

Time is taken by the stone to hit the ground 3.75 seconds

Distance moved by the stone = 16.06 ×3.75 m

= 60.3 m

Thus, the stone land 60.3 m away from the foot of the cliff

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Your question seems incomplete the complete question with the figure is

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 40.09 with the face of the cliff,as shown in the figure. The stone hits the ground 3.75 $ after being thrown and feels no appreciable air resistance as it falls How far from the foot of the cliff does the stone land?

(A) 6 m

(B) 25.0 m

(C)93.8 m

(D) 60.3 m