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 m/s. The first snowball is thrown at an angle of 52◦ 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/s2.
To solve this problem, we can use the kinematic equations of motion to analyze the horizontal and vertical components of the snowball's motion.
Let's break down the motion of the first snowball into its horizontal and vertical components:
1. Horizontal motion:
Since the snowball is thrown over level ground, there is no acceleration acting horizontally. Therefore, the horizontal component of the initial velocity, Vx1, remains constant throughout the motion.
Vx1 = V * cosθ1
where V is the initial speed of the snowball (32 m/s) and θ1 is the angle at which it is thrown (52° above the horizontal).
2. Vertical motion:
The vertical motion of the snowball is subject to acceleration due to gravity. We can use the kinematic equation to calculate the time of flight, tf, for the first snowball:
0 = Vy1 - gt
where Vy1 is the vertical component of the initial velocity and g is the acceleration due to gravity (-9.8 m/s^2).
Rearranging the equation, we get:
Vy1 = gt
Now, we can plug in the values:
Vy1 = (9.8 m/s^2)(tf)
Next, we can calculate the vertical displacement, Δy, of the first snowball using another kinematic equation:
Δy = Vy1 * tf + (1/2) * g * tf^2
At the peak of the trajectory, the vertical displacement is zero since the starting and ending heights are the same.
Now, let's analyze the second snowball thrown at a lower angle. We want it to hit the same point as the first snowball, so the horizontal displacement, Δx, of the second snowball should be the same as the first snowball.
1. Horizontal displacement (Δx):
Given that both snowballs hit the same point, the horizontal displacements of both snowballs are equal because they cover the same horizontal distance.
Δx = Vx1 * tf
2. Vertical motion:
The second snowball's vertical motion is also subject to acceleration due to gravity. We want to determine the angle θ2 at which the second snowball should be thrown to hit the same point as the first snowball.
We can calculate the vertical displacement, Δy, of the second snowball using the same kinematic equation as before:
Δy = Vy2 * tf + (1/2) * g * tf^2
Here, we need to find the vertical component of the initial velocity, Vy2, using the initial speed (32 m/s) and the angle θ2.
Now, the key is to equate the horizontal displacements and vertical displacements of the two snowballs since they both hit the same point:
Δx = Δy
Substituting the respective equations, we get:
Vx1 * tf = Vy2 * tf + (1/2) * g * tf^2
Now, we can solve this equation to find the angle θ2 at which the second snowball should be thrown in order to hit the same point as the first snowball.