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 13.4 m/s. The first
snowball is thrown at an angle of 53◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of ◦
To solve this problem, we can use the principles of projectile motion and set up two equations: one for the vertical motion of the first snowball and one for the horizontal motion of both snowballs.
Let's start by analyzing the vertical motion of the first snowball. We can break down its initial velocity into horizontal and vertical components. The vertical component can be determined using trigonometry:
Vertical component = initial speed * sin(angle)
In this case, the initial speed is 13.4 m/s and the angle is 53°. Plugging in these values, we can calculate the vertical component of the first snowball's velocity:
Vertical component = 13.4 m/s * sin(53°)
Vertical component ≈ 10.48 m/s
Next, let's analyze the horizontal motion of both snowballs. The horizontal component of velocity remains constant throughout the entire motion and is given by:
Horizontal component = initial speed * cos(angle)
Again, the initial speed is 13.4 m/s. For the first snowball, the angle is 53°, but for the second snowball, we're looking for the angle that will make it hit the same point as the first snowball. Let's call this angle θ.
Thus, the horizontal component of the second snowball's velocity can be written as:
Horizontal component (second snowball) = 13.4 m/s * cos(θ)
Now, we can set up an equation to determine the time it takes for both snowballs to reach the target point. Since the first snowball is thrown at a high angle, it will spend more time in the air compared to the second snowball thrown at a low angle. The equation for time can be derived from the vertical motion:
Time = (2 * vertical velocity) / acceleration due to gravity
In this case, the vertical velocity is 10.48 m/s, and the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the vertical motion). Plugging in these values, we can calculate the time of flight for the first snowball:
Time (first snowball) = (2 * 10.48 m/s) / 9.8 m/s^2
Time (first snowball) ≈ 2.08 s
Since we want the second snowball to hit the same point at the same time, its time of flight should also be 2.08 seconds. We can set up an equation for the horizontal motion of the second snowball:
Horizontal distance = (Horizontal component) * (Time)
For the first snowball, the horizontal distance is:
Horizontal distance (first snowball) = 13.4 m/s * cos(53°) * 2.08 s
For the second snowball, the horizontal distance is:
Horizontal distance (second snowball) = 13.4 m/s * cos(θ) * 2.08 s
Since the two snowballs hit the same point, the horizontal distances should be equal:
13.4 m/s * cos(53°) * 2.08 s = 13.4 m/s * cos(θ) * 2.08 s
Now, we can solve this equation to find θ, the angle at which the second snowball should be thrown. Canceling out the common terms, we get:
cos(53°) = cos(θ)
To find the angle θ, we can take the inverse cosine of both sides of the equation:
θ = cos^(-1)(cos(53°))
Evaluating this expression, we find:
θ ≈ 53°
Therefore, the angle at which you should throw the second snowball to make it hit the same point as the first snowball is approximately 53°.
Answer: θ ≈ 53°
answer: 19 degrees
explanation:
when i write v0 it means v initial
distance traveled by snowball 1
d1 = v0t1 = (v*cos(theta1)) * [ (2v0*sin(theta1)) / (g) ]
since sin(2(theta)) is 2sin(theta)cos(theta)
d1 = ((v0)^2 / g)*sin2(theta1)
for the same reason,
d2 = ((v0)^2 / g)*sin2(theta2)
set d1 equal to d2 and solve for theta2 with calculator or with trig identities