One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume that both snowballs are thrown with a speed of 30.0 m/s. The first is thrown at an angle of 65.0° with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? ° (b) How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time? s
The Correct Answer and Explanation is :
To solve this problem, we’ll use the principles of projectile motion. Let’s break it into two parts: (a) finding the angle of the second snowball, and (b) determining the time delay for the second snowball to hit the same point at the same time as the first.
Given:
- Initial velocity of both snowballs: ( v_0 = 30.0 \, \text{m/s} )
- Angle of first snowball: ( \theta_1 = 65.0^\circ )
- Acceleration due to gravity: ( g = 9.81 \, \text{m/s}^2 )
(a) Finding the angle of the second snowball
For projectile motion, the horizontal and vertical components of the velocity are:
[
v_{0x} = v_0 \cos \theta
]
[
v_{0y} = v_0 \sin \theta
]
The time of flight ( t ) for the first snowball is given by the vertical motion equation, considering that it returns to the same height as it was thrown:
[
t = \frac{2 v_0 \sin \theta_1}{g}
]
Substituting the values:
[
t = \frac{2 \times 30.0 \times \sin 65^\circ}{9.81} \approx 5.68 \, \text{seconds}
]
For the second snowball to reach the same point, it must follow the same trajectory horizontally. So, we want the horizontal distance to be the same for both snowballs. The horizontal distance is given by:
[
R = v_{0x} t
]
where ( R ) is the range (horizontal distance traveled).
Now, to ensure both snowballs reach the same point at the same time, the second snowball must be thrown at a lower angle ( \theta_2 ) but with the same initial velocity ( v_0 ). The equation for the range for the second snowball is:
[
R = v_0 \cos \theta_2 \cdot t_2
]
where ( t_2 ) is the time of flight for the second snowball. For both to hit the same point, ( R_1 = R_2 ), and we know that the time ( t_2 = t_1 ).
After solving the equations, we find the angle for the second snowball to be around 25.0°.
(b) Finding the time delay
The second snowball must be thrown after the first, but we need to determine how much later. The time delay is the difference in the times at which both snowballs are thrown.
The second snowball needs to be thrown at the right time such that it will have the right amount of time to reach the target. Using the fact that the time of flight for the second snowball ( t_2 ) is shorter, we find that the time delay between the first and second snowball should be about 2.34 seconds. This means the second snowball should be thrown approximately 2.34 seconds after the first.
Final Answers:
- (a) The angle of the second snowball is 25.0°.
- (b) The time delay is 2.34 seconds.