Question

A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall, 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 the one with the same initial speed 14.1 m/s. The first snow ball is thrown at an angle of 52° above the horizontal. At what angle Sure you throw the second snowball to make ahead the same point as the first? Acceleration of gravity is 9.8 m/s^2. Answer in units of °.
b) How many seconds after the first snowball should be through the second so that they arrive on target at the same time? Answer in units of s.

Answer 1

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

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:

$V_{o}=14.1 m/s$ is the initial speed  of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ 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)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$   (9)

$x=19.684 m$   (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)

$2\theta_{2}=75.99\°$  

$\theta_{2}=37.99\° \approx 38\°$  (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$.

b) Time difference between both snowballs

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

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

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$   (16)

$t_{1}=2.267 s$   (17)

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

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$   (18)

$t_{2}=1.771 s$   (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$   (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$   (21)

$t=2.267 s - 1.771 s$  

Finally:

$t=0.495 s$