During a snowball �ght, your opponent distracts you by throwing a snowball at you with a high arc. She throws snowballs with a speed of 26.5 m/s and the �first is thrown at 70.0 degrees. As you are watching the first snowball, she throws a second at a lower angle.
(a) If both snowballs cover the same horizontal distance, at what angle (degrees) should
the second be thrown?
(b) If the second snow ball is going to hit you while your attention is still focused on the �first snowball, how soon (s) after the first snowball is thrown must the second be thrown?
In other words, what is the delay between the two throws that will have both arriving at
the same time?
20
a. 20 degrees. Both angles have to equal 90 degrees total.
To solve this problem, we need to analyze the motion of the two snowballs.
(a) To find the angle at which the second snowball should be thrown, we can use the principle of projectile motion. We know that both snowballs cover the same horizontal distance. Since the horizontal distance is the same, the horizontal velocity component will also be the same for both snowballs.
Let's assume that the horizontal distance covered by each snowball is "d".
For the first snowball:
Initial velocity (v1) = 26.5 m/s
Launch angle (θ1) = 70.0 degrees
For the second snowball:
Initial velocity (v2) = ?
Launch angle (θ2) = angle we want to find
Using the principle of projectile motion, we can write the equations for the horizontal component and the vertical component of the snowball's motion.
Horizontal component:
For both snowballs, the horizontal velocity (Vx) remains the same because the horizontal distance is the same.
Vx = v1 * cos(θ1) = v2 * cos(θ2)
Vertical component:
For the first snowball:
Vy1 = v1 * sin(θ1)
For the second snowball:
Vy2 = v2 * sin(θ2)
Since the two snowballs hit the ground at the same time, we can set the time of flight (t1) for the first snowball equal to the time of flight (t2) for the second snowball.
t1 = t2
We can find the time of flight using the vertical component equation:
t = 2 * (Vy / g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the first snowball:
t1 = 2 * (Vy1 / g)
For the second snowball:
t2 = 2 * (Vy2 / g)
Since t1 = t2, we can equate the expressions for t1 and t2:
2 * (Vy1 / g) = 2 * (Vy2 / g)
Simplifying, we get:
Vy1 = Vy2
Substituting the equations for Vy1 and Vy2, we get:
v1 * sin(θ1) = v2 * sin(θ2)
We know the values for v1, θ1, and v2, so we can substitute those in and solve for θ2.
(b) To find the time delay between the two throws that will have both snowballs arriving at the same time, we need to find the difference between the flight times of the two snowballs.
Let's assume the time delay between the two throws is "t".
The total flight time for the first snowball is 2 * t1, and for the second snowball, it is 2 * t2.
Since we want both snowballs to arrive at the same time, we can set the flight times equal to each other:
2 * t1 = 2 * t2
Since we know the expressions for t1 and t2, we can substitute those in and solve for t.