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 12.6 m/s. The first
snowball is thrown at an angle of 60

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

You use this service a lot. How about showing your own work and/or thoughts for a change??

In this case, they are asking for wnat other angle will cause the snowball to reach the same spot, when thrown at the same velocity, Vo.

Learn (or derive) this formula:

Range = (Vo^2/g)*sin(2A)

Since the range is proportional to the sine of twice the launch angle A, the other angle is 30 degrees. The sines of 60 and 120 degrees are the same. The time of flight will be less when thrown at 30 degrees.

To solve this problem, we can use the kinematic equations of projectile motion.

We are given the initial speed of both snowballs, which is 12.6 m/s. Let's denote this as v0.

For the first snowball:
Initial speed (v0) = 12.6 m/s
Launch angle (θ1) = 60°

For the second snowball:
Initial speed (v0) = 12.6 m/s
Launch angle (θ2) = ? (what we need to find)

The horizontal and vertical components of the velocities for both snowballs are:

First snowball:
Vx1 = v0 * cos(θ1)
Vy1 = v0 * sin(θ1)

Second snowball:
Vx2 = v0 * cos(θ2)
Vy2 = v0 * sin(θ2)

Since we want the two snowballs to hit the same point at the same time, we can equate their horizontal and vertical positions:

For the horizontal position:
x1 = x2 (They travel the same horizontal distance)
To get x1 and x2, we need to determine the time of flight for each snowball.
The time of flight for a projectile can be calculated as:
t = (2 * Vy) / g (where g is the acceleration due to gravity)

For the vertical position:
y1 = y2 (They hit the same point vertically)
Similarly, we need to determine the maximum heights reached by each snowball.
The maximum height can be calculated as:
y_max = (Vy^2) / (2 * g)

Let's calculate the values for the first snowball, using the given values:
Vx1 = v0 * cos(θ1) = 12.6 * cos(60°)
Vy1 = v0 * sin(θ1) = 12.6 * sin(60°)

Using the time of flight equation:
t1 = (2 * Vy1) / g = (2 * 12.6 * sin(60°)) / 9.8

Using the maximum height equation:
y_max1 = (Vy1^2) / (2 * g) = (12.6 * sin(60°))^2 / (2 * 9.8)

Now, let's calculate the values for the second snowball, where we need to find θ2:
Vx2 = v0 * cos(θ2) = 12.6 * cos(θ2)
Vy2 = v0 * sin(θ2) = 12.6 * sin(θ2)

Using the time of flight equation:
t2 = (2 * Vy2) / g = (2 * 12.6 * sin(θ2)) / 9.8

Using the maximum height equation:
y_max2 = (Vy2^2) / (2 * g) = (12.6 * sin(θ2))^2 / (2 * 9.8)

Since we want the two snowballs to hit the same point, we need to equate the horizontal positions and the maximum heights:

x1 = x2
y_max1 = y_max2

By setting up and solving the equations, we can find θ2.

To solve this problem, we can use the equations of motion to find the time of flight for each snowball.

Let's start by finding the time of flight for the first snowball.

The initial velocity of the first snowball can be split into its horizontal and vertical components.
The vertical component of the velocity is given by V_y = V_initial * sin(theta), where V_initial is the initial speed (12.6 m/s) and theta is the angle of 60 degrees.

Using the equation of motion for vertical motion, we have:
V_y = V_initial * sin(theta)
0 = V_initial * sin(theta) - g * t
t = V_initial * sin(theta) / g

Now, let's find the time of flight for the second snowball.

The initial velocity of the second snowball can also be split into its horizontal and vertical components.
The vertical component of the velocity is given by V_y = V_initial * sin(theta'), where V_initial is the initial speed (12.6 m/s) and theta' is the angle we want to find.

Using the equation of motion for vertical motion, we have:
V_y = V_initial * sin(theta')
0 = V_initial * sin(theta') - g * t'
t' = V_initial * sin(theta') / g

Since we want both snowballs to hit the same point simultaneously, the time of flight for both snowballs should be the same. Therefore, t = t'.

Now, let's solve for theta' by equating the expressions for t and t':

V_initial * sin(theta) / g = V_initial * sin(theta') / g

Canceling out the common terms, we get:

sin(theta) = sin(theta')

To find the angle theta', we can take the inverse sine (also known as arcsine) of both sides:

theta' = arcsin(sin(theta))

Now we can substitute the given value of theta (60 degrees) into the equation:

theta' = arcsin(sin(60))

Calculating this expression, we find:

theta' ≈ 60 degrees

Therefore, to make the second snowball hit the same point as the first snowball, you should throw it at an angle of approximately 60 degrees above the horizontal.