Physics, 12.10.2019 01:20 SoccerHalo

One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
while your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
assume both snowballs are thrown with
the same initial speed 11.1 m/s. the first
snowball is thrown at an angle of 67◦
above
the horizontal. at what angle should you
throw the second snowball to make it hit the
same point as the first? the acceleration of
gravity is 9.8 m/s
2
.
answer in units of ◦
.
007 (part 2 of 2) 10.0 points
how many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
answer in units of s.

Answers: 1

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Answers

Answer from: NijaNij

Answers:

a) $\theta_{2}=23\°$

b) t=1.199 s

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: x-component and y-component. Being their main equations as follows for both snowballs:

Snowball 1:

x-component:

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=11.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=67\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

y-component:

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

y=0 is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

Snowball 2:

x-component:

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

y-component:

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

a) Angle for snowball 2

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))$ (9)

x=9.043 m (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=45.99\°$

$\theta_{2}=22.99\° \approx 23\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .