# Skill Set 03.2

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Main Page > IB Physics Skill Sets > Skill Set 03.2

### 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

### 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

### 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

### 4. A ball is launched at 43.2 m/s at an angle of 25.2o above horizontal on a level field. How far does the ball go before striking the ground?

The first thing to do is to split the velocity into x and y components. To do this, we do the following:

43.2cos(25.2) = 39.089 m/s
43.2sin(25.2) = 18.394 m/s

These are our initial velocities, but since horizontal acceleration is 0, our final horizontal velocity is the same as our initial. And for the y component, since the ball begins and ends on the same level ground, the final velocity will be the negative of the initial.

For problem 4, our suvat table will now look like this:

H V

s =
u = 39.089 m/s
v = 39.089 m/s
a = 0
t =

s = 0
u = 18.394 m/s
v = -18.394 m/s
a = -9.81 m/s/s
t =

Since we need to find horizontal displacement (s), from s = ut + 1/2at2, we will first need to find t.

It is a simple matter to use v = u + at (if you remember it) to find time in the vertical direction:
-18.394 = 18.394 + (-9.81 m/s2)t
t = 3.7500 s

Now using s = ut + 1/2at2 with a = 0,

s = (39.089 m/s)(3.75 s)
s = 146.581 = 147 m

### 5. A ball is launched at 43.2 m/s at an angle of 25.2o above horizontal on a level field. There is a very tall wall 130. m away. With what velocity does the ball strike the wall? (Express your answer as an angle-magnitude vector – draw a picture)

We will need to use a new suvat table for this problem, because factors are affected with the shortened distance the ball can now travel. We will, however, use the same initial velocities we received from question 4.

H V

s = 130. m
u = 39.089 m/s
v = 39.089 m/s
a = 0
t =

s =
u = 18.394 m/s
v =
a = -9.81 m/s/s
t =

In order to get a final velocity vector, we will need both the x and y velocities. Since we have the x, our next step is to solve for the final velocity of y. We can do that by using v = u + at, but we first need t, which we will receive by doing the horizontal s = ut + 1/2at2 with a being 0.

s = ut
130 m = (39.089 m/s)(t)
t = 3.3258 s

Now using the vertical form of v = u + at:

v = 18.394 m/s + (-9.81 m/s2)(3.3258 s)
v = -14.232 m/s

Now that we have both of our final velocities, we can convert them into an angle magnitude vector.

a2 + b2 = c2
39.0892 + -14.2322 = c2
c = 41.6 m/s
tan-1 = -20.0o
41.6 m/s @ 20o below the horizontal