Question

006 (part 1 of 2) 10.0 points

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 32.2 m/s. The first
snowball is thrown at an angle of 51° 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?.
Answer in units of º.
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.

Answer 1

Answers:

a)
`\theta_(2)=39\°`

b)
`t=0.972 s`

Step-by-step 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)=32.2 m/s` is the initial speed of snowball 1 (and snowball 2, as well)

`\theta_(1)=51\°` 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)+(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)+(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)+(gt_(1)^(2))/(2)` (5)

`t_(1)=-(2V_(o)sin\theta_(1))/(g)` (6)

Substituting (6) in (1):

`x=V_(o)cos\theta_(1)(-(2V_(o)sin\theta_(1))/(g))` (7)

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

`x=-(V_(o)^(2))/(g) sin(2\theta_(1))` (8)

`x=-((32.2 m/s)^(2))/(-9.8 m/s^(2)) sin(2(51\°))` (9)

`x=103.488 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)=-(2V_(o)sin\theta_(2))/(g)` (11)

`x=V_(o)cos\theta_(2) (-(2V_(o)sin\theta_(2))/(g))` (12)

Rewritting (12):

`x=-(V_(o)^(2))/(g) sin(2\theta_(2))` (13)

Finding
`\theta_(2)`:

`2\theta_(2)=sin^(-1)((-xg)/(V_(o)^(2)))` (14)

`2\theta_(2)=77.99\°`

`\theta_(2)=38.99\° \approx 39\°` (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)=-(2V_(o)sin\theta_(1))/(g)`

`t_(1)=-(2(32.2 m/s)sin(51\°))/(-9.8m/s^(2))` (16)

`t_(1)=5.106 s` (17)

`t_(2)=-(2V_(o)sin\theta_(2))/(g)`

`t_(2)=-(2(32.2 m/s)sin(39\°))/(-9.8m/s^(2))` (18)

`t_(2)=4.134 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=5.106 s - 4.134 s`

Finally:

`t=0.972 s`