View source for Skill Set 03.2

[[Main Page]] > [[IB Physics Skill Sets]] > Skill Set 03.2
<br />
<br />
===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? ===
<blockquote>
To start off, our suvat table for problems 1-3 will originally look like this:<br>
{| border="1" cellpadding="10"
|+
!width="150"|H
!width="150"|V
|-
|
s = <br>
u = 17.3 m/s<br>
v = 17.3 m/s<br>
a = 0<br>
t = 1.56 s<br>

|
s = <br>
u = 0 (cliff)<br>
v = <br>
a = -9.81 m/s/s<br>
t = 1.56 s<br>

|-
|}
Using s = ut + 1/2at<sup>2</sup>:<br><br>
s = 0 + 1/2(9.81)(1.56)<sup>2</sup> = '''11.936 m'''
</blockquote>
[[#top | Table of Contents]]
----

===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?===
<blockquote>
We will again use s = ut + 1/2at<sup>2</sup>, but since horizontal acceleration = 0, the equation becomes s = ut.<br><br>

s = ut = (17.3)(1.56) = '''26.988 m'''
</blockquote>
[[#top | Table of Contents]]
----

===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?===
<blockquote>
So in the vertical direction you can use v = u + at to solve for the final vertical velocity.<br><br>

v = 0 + (-9.81 m/s<sup>2</sup>)(1.56 s)<br>
v = -15.3036 m/s<br><br>

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.<br><br>

17.3<sup>2</sup> + -15.3<sup>2</sup> = c<sup>2</sup><br>
c = '''23.1 m/s'''
</blockquote>
[[#top | Table of Contents]]
----

===4. A ball is launched at 43.2 m/s at an angle of 25.2<sup>o</sup> above horizontal on a level field. How far does the ball go before striking the ground?===
<blockquote>
The first thing to do is to split the velocity into x and y components. To do this, we do the following:<br><br>

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

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.<br><br><br>


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

{| border="1" cellpadding="10"
|+
!width="150"|H
!width="150"|V
|-
|
s = <br>
u = 39.089 m/s<br>
v = 39.089 m/s<br>
a = 0<br>
t = <br>

|
s = 0<br>
u = 18.394 m/s<br>
v = -18.394 m/s<br>
a = -9.81 m/s/s<br>
t = <br>

|-
|}
</blockquote>

Since we need to find horizontal displacement (s), from s = ut + 1/2at<sup>2</sup>, we will first need to find t.<br><br>

It is a simple matter to use v = u + at (if you remember it) to find time in the vertical direction:<br>
-18.394 = 18.394 + (-9.81 m/s<sup>2</sup>)t<br>
t = 3.7500 s<br><br>

Now using s = ut + 1/2at<sup>2</sup> with a = 0,<br><br>

s = (39.089 m/s)(3.75 s)<br>
s = 146.581 = '''147 m'''

[[#top | Table of Contents]]
----

===5. Question===
<blockquote>
Solution goes here
</blockquote>
[[#top | Table of Contents]]
----