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.
The acceleration of gravity is 9.8 m/s2 .
Assume both snowballs are thrown with
the same initial speed 35.2 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?
Answer in units of �.
We can solve this problem by breaking down the motion of the two snowballs into horizontal and vertical components.
Let's focus on the first snowball thrown at a 51° angle. The initial velocities for the first snowball in each direction are:
Vx1 = 35.2 * cos(51°)
Vy1 = 35.2 * sin(51°)
Now, we need to find the time it takes for the first snowball to reach its target. To do this, we will use the following equation for the vertical motion (since the second snowball will have the same time in reaching the target):
y = Vy1 * t - (1/2) * g * t^2
Since the snowball is thrown and lands at the same height, the vertical displacement (y) will be zero:
0 = Vy1 * t - (1/2) * g * t^2
Rearranging to solve for t:
t = (2 * Vy1) / g
Now, let's find the horizontal distance (x) traveled by the first snowball:
x = Vx1 * t
The second snowball is thrown at a lower angle θ. The initial velocities for the second snowball in each direction are:
Vx2 = 35.2 * cos(θ)
Vy2 = 35.2 * sin(θ)
The horizontal distance traveled by the second snowball is the same as the first snowball:
x = Vx2 * t
Since we know that t is the same for both snowballs, we can set the two horizontal distance equations equal to each other:
Vx1 * t = Vx2 * t
Canceling the t on both sides:
Vx1 = Vx2
We can now substitute the values of Vx1 and Vx2 in terms of the two angles:
35.2 * cos(51°) = 35.2 * cos(θ)
Since the initial speed is the same for both snowballs, we can divide both sides by 35.2 to simplify:
cos(51°) = cos(θ)
Now, to solve for θ:
θ = arccos(cos(51°))
θ ≈ 39°
Therefore, you should throw the second snowball at a 39° angle to make it hit the same point as the first snowball.
To solve this problem, we can use the kinematic equations of motion. Let's start by finding the time it takes for the first snowball to hit the ground.
We have the initial velocity (v₀ = 35.2 m/s) and the launch angle (θ₁ = 51 degrees) of the first snowball. First, we need to split the initial velocity into its horizontal and vertical components:
vx = v₀ * cos(θ₁)
vy = v₀ * sin(θ₁)
Since the snowball is thrown over level ground, the vertical component of velocity at the highest point of its trajectory is zero. We can use this fact to find the time it takes for the first snowball to reach the highest point. In this case, the vertical component of velocity can be determined using the following equation:
vy = v₀ * sin(θ) - g * t₁
0 = v₀ * sin(θ₁) - g * t₁
Solving for t₁ gives us:
t₁ = v₀ * sin(θ₁) / g
Now, let's determine the time it takes for the second snowball to reach the same point. Since both snowballs hit the ground at the same time, the total time of flight for the second snowball must be equal to t₁.
To find the launch angle of the second snowball (θ₂), we can use the following equation:
y = v₀ * sin(θ₂) * t₂ - (1/2) * g * t₂²
Here, y is the height at which the snowball hits the ground, which is the same as the height of the first snowball. We know that t₂ must be equal to t₁, so we can substitute t₁ into the equation:
y = v₀ * sin(θ₂) * t₁ - (1/2) * g * t₁²
Now we can solve for θ₂. Rearranging the equation:
0 = (1/2) * g * t₁² - v₀ * sin(θ₂) * t₁ + y
Since the snowballs hit the same point, y = 0. Substituting y = 0 and rearranging the equation:
(1/2) * g * t₁² = v₀ * sin(θ₂) * t₁
Now we can solve for θ₂:
sin(θ₂) = (1/2) * g * t₁ / v₀
θ₂ = arcsin((1/2) * g * t₁ / v₀)
Substituting the known values:
θ₂ = arcsin((1/2) * 9.8 * t₁ / 35.2)
Now we can substitute t₁ = v₀ * sin(θ₁) / g:
θ₂ = arcsin((1/2) * 9.8 * (v₀ * sin(θ₁) / g) / 35.2)
Simplifying:
θ₂ = arcsin(0.5 * sin(θ₁))
Now we can calculate θ₂:
θ₂ = arcsin(0.5 * sin(51))
Using a calculator, we find that:
θ₂ ≈ 15.5 degrees
Therefore, the angle at which you should throw the second snowball to hit the same point as the first is approximately 15.5 degrees.
To find the angle at which the second snowball should be thrown, we can use the range formula for projectile motion.
The range formula is given by:
Range = (V^2 * sin(2θ)) / g
Where:
- Range is the horizontal distance traveled by the projectile
- V is the initial speed of the projectile
- θ is the launch angle of the projectile
- g is the acceleration due to gravity
Both snowballs need to hit the same point, so their ranges should be equal.
For the first snowball:
Range1 = (V^2 * sin(2θ1)) / g
For the second snowball:
Range2 = (V^2 * sin(2θ2)) / g
Since both ranges are equal, we can set them equal to each other and solve for θ2:
(V^2 * sin(2θ1)) / g = (V^2 * sin(2θ2)) / g
Simplifying:
sin(2θ1) = sin(2θ2)
Using the double angle formula for sine:
2sin(θ1)cos(θ1) = 2sin(θ2)cos(θ2)
Dividing both sides by 2:
sin(θ1)cos(θ1) = sin(θ2)cos(θ2)
Now, we can use the identity sin(2θ) = 2sin(θ)cos(θ) to rewrite the equation:
sin(θ1)cos(θ1) = sin(θ2)(1 - sin^2(θ2))
Expanding and simplifying:
sin(θ1)cos(θ1) = sin(θ2) - sin^3(θ2)
Rearranging the equation:
sin^3(θ2) + sin(θ2) - sin(θ1)cos(θ1) = 0
Now, we can solve this equation to find θ2 using numerical methods such as graphing or using a calculator.
Given the values:
V = 35.2 m/s
θ1 = 51 degrees
g = 9.8 m/s^2
We substitute the values into the equation and solve for θ2.