Skill Set 03.2

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1. A ball rolls off the edge of a cliff. The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds. How high is the cliff?

To start off, our suvat table for problems 1-3 will originally look like this:

H	V
s = u = 17.3 m/s v = 17.3 m/s a = 0 t = 1.56 s	s = u = 0 (cliff) v = a = -9.81 m/s/s t = 1.56 s

Using s = ut + 1/2at²:

s = 0 + 1/2(9.81)(1.56)² = 11.936 m

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2. A ball rolls off the edge of a cliff. The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds. How far from the base of the cliff does the ball land?

We will again use s = ut + 1/2at², but since horizontal acceleration = 0, the equation becomes s = ut.

s = ut = (17.3)(1.56) = 26.988 m

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3. A ball rolls off the edge of a cliff. The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds. What is the speed of impact?

So in the vertical direction you can use v = u + at to solve for the final vertical velocity.

v = 0 + (-9.81 m/s²)(1.56 s)
v = -15.3036 m/s

Since the acceleration in the horizontal direction is 0, the initial velocity of 17.3 m/s is also the final velocity. With both our horizontal and vertical final velocities, we can find the magnitude of the impact by using the Pythagorean Theorem.

17.3² + -15.3² = c²
c = 23.1 m/s

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4. Question

Solution goes here

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5. Question

Solution goes here

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