In this problem, we are asked to find the angle at which the second snowball should be thrown to hit the same point as the first snowball, given that both snowballs have the same initial speed and that the starting and ending heights are the same. This is a classic projectile motion problem.

Step-by-Step Solution:

Let’s break this down into steps:

1. Initial Data:
   • Initial speed of both snowballs: $v_0 = 28.9 \, \text{m/s}$
   • Angle of the first snowball: $\theta_1 = 74^\circ$
   • Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$
   • The second snowball is thrown at some unknown angle $\theta_2$.
   • Both snowballs should land at the same point at the same time.

Horizontal Motion:

The horizontal range $R$ of a projectile is given by: $R = \frac{v_0^2 \sin(2\theta)}{g}$ where $v_0$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

Since both snowballs must travel the same horizontal distance, their ranges must be the same. This means that the horizontal ranges of both snowballs must be equal.

Let’s calculate the range $R_1$ for the first snowball using $\theta_1 = 74^\circ$: $R_1 = \frac{v_0^2 \sin(2 \times 74^\circ)}{g} = \frac{(28.9)^2 \sin(148^\circ)}{9.8}$

Using $\sin(148^\circ) \approx 0.5299$: $R_1 = \frac{(28.9)^2 \times 0.5299}{9.8} = \frac{835.21 \times 0.5299}{9.8} \approx \frac{442.65}{9.8} \approx 45.16 \, \text{m}$

Thus, the range for the first snowball is approximately $R_1 = 45.16 \, \text{m}$.

Time of Flight:

The time of flight $T$ for a projectile is given by: $T = \frac{2v_0 \sin(\theta)}{g}$

For the first snowball: $T_1 = \frac{2 \times 28.9 \times \sin(74^\circ)}{9.8}$

Using $\sin(74^\circ) \approx 0.9613$: $T_1 = \frac{2 \times 28.9 \times 0.9613}{9.8} = \frac{55.57}{9.8} \approx 5.67 \, \text{seconds}$

So, the first snowball is in the air for about $5.67 \, \text{seconds}$.

Time of Flight for the Second Snowball:

Since both snowballs must hit the same point at the same time, the second snowball must also have a time of flight of $T_2 = 5.67 \, \text{seconds}$.

Using the formula for time of flight: $T_2 = \frac{2v_0 \sin(\theta_2)}{g}$ we can solve for $\theta_2$: $5.67 = \frac{2 \times 28.9 \times \sin(\theta_2)}{9.8}$ $5.67 \times 9.8 = 2 \times 28.9 \times \sin(\theta_2)$ $55.57 = 57.8 \times \sin(\theta_2)$ $\sin(\theta_2) = \frac{55.57}{57.8} \approx 0.9613$ $\theta_2 = \arcsin(0.9613) \approx 74^\circ$

However, this is the same angle as the first snowball. To hit the same point with a lower angle, we use the fact that there are two angles that give the same range for the same initial speed: one high angle and one low angle.

The complementary angle that gives the same range is: $\theta_2 = 90^\circ - 74^\circ = 16^\circ$

Final Answer:

The second snowball should be thrown at an angle of 16 degrees to make it hit the same point as the first snowball.

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Would you like more details, or do you have any questions? Here are five related questions to deepen understanding:

1. How does the time of flight change if the initial speed is increased?
2. What happens if the first snowball is thrown at a higher angle, say 80°?
3. Can you calculate the range for the second snowball thrown at 16°?
4. How would air resistance affect the time of flight of both snowballs?
5. What would the trajectory of the snowballs look like if visualized?

Tip: When solving projectile motion problems, it's helpful to treat horizontal and vertical motions separately and use symmetry whenever possible.