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 32.1 m/s. The first snowball is thrown at an angle of 57◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? Note the starting and ending heights are the same. The acceleration of gravity is 9.8 m/s 2 . (Part 2) 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.
The Correct Answer and Explanation is :
Part 1: Determining the angle of the second snowball
We are given the following information for the first snowball:
- Initial speed: ( v_0 = 32.1 \, \text{m/s} )
- Launch angle: ( \theta_1 = 57^\circ )
- Acceleration due to gravity: ( g = 9.8 \, \text{m/s}^2 )
We need to find the angle ( \theta_2 ) for the second snowball to ensure it hits the same point at the same time. The horizontal and vertical components of the motion must be the same for both snowballs, and the total time of flight will be equal.
Step 1: Determine the time of flight for the first snowball
We start by determining the time it takes for the first snowball to reach the target. The formula for the time of flight ( t_1 ) of a projectile is given by:
[
t_1 = \frac{2 v_0 \sin(\theta_1)}{g}
]
Substitute the known values:
[
t_1 = \frac{2 \times 32.1 \times \sin(57^\circ)}{9.8}
]
[
t_1 \approx \frac{2 \times 32.1 \times 0.8387}{9.8}
]
[
t_1 \approx \frac{53.8}{9.8} \approx 5.5 \, \text{seconds}
]
Step 2: Horizontal distance of the first snowball
Next, we calculate the horizontal distance traveled by the first snowball. The formula for horizontal displacement ( R_1 ) is:
[
R_1 = v_0 \cos(\theta_1) \times t_1
]
Substitute the known values:
[
R_1 = 32.1 \times \cos(57^\circ) \times 5.5
]
[
R_1 \approx 32.1 \times 0.5446 \times 5.5 \approx 95.5 \, \text{m}
]
Step 3: Angle for the second snowball
Now we use the same principles for the second snowball, which has the same initial speed but will be thrown at a different angle ( \theta_2 ). To hit the same point at the same time, the horizontal distance and time of flight must be the same.
We already know ( t_2 = t_1 \approx 5.5 \, \text{seconds} ), so we can use the horizontal displacement equation for the second snowball:
[
R_2 = v_0 \cos(\theta_2) \times t_2
]
Since ( R_1 = R_2 ), we have:
[
v_0 \cos(\theta_1) \times t_1 = v_0 \cos(\theta_2) \times t_2
]
Simplifying and solving for ( \theta_2 ):
[
\cos(\theta_1) = \cos(\theta_2)
]
Since the angles must satisfy this condition, we find that:
[
\theta_2 = 90^\circ – \theta_1
]
Thus,
[
\theta_2 = 90^\circ – 57^\circ = 33^\circ
]
Part 2: Timing the second snowball
We already know the time of flight for both snowballs is the same. Therefore, the second snowball should be thrown at the same time as the first one to ensure they both arrive at the target at the same time.
The time to throw the second snowball after the first snowball is 0 seconds.
Summary
- Angle for the second snowball: ( \theta_2 = 33^\circ )
- Time delay: 0 seconds (the second snowball should be thrown at the same time as the first).
Image
I will generate an image showing the projectile motion of both snowballs, indicating their trajectories with the given launch angles.
Here is the diagram illustrating the projectile motion of both snowballs. The first snowball is launched at a 57° angle, and the second snowball is launched at a 33° angle. Both hit the ground at the same point, demonstrating the principle of projectile motion for different launch angles with the same initial speed.
