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 17.9 m/s. The first
snowball is thrown at an angle of 50◦
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 determine the angle at which the second snowball should be thrown, we can use the principle of projectile motion.
First, let's analyze the motion of the first snowball. It is thrown at an angle of 50° above the horizontal, with an initial speed of 17.9 m/s.
We can break down the initial velocity of the first snowball into two components: the horizontal component (Vx) and the vertical component (Vy).
Vx = V * cos(theta)
Vy = V * sin(theta)
where V is the initial speed (17.9 m/s) and theta is the angle of projection (50°).
For the first snowball, the time it takes to reach the same point will be the same as for the second snowball. Let's call this time "t".
Now, let's analyze the motion of the second snowball. We need to find the angle at which it should be thrown.
To find the angle, let's break down the velocity of the second snowball into horizontal and vertical components similar to the first snowball.
Vx2 = V2 * cos(theta2)
Vy2 = V2 * sin(theta2)
The initial speed of the second snowball is also 17.9 m/s. We can use the principle of projectile motion to relate the two velocities at the same time "t".
Since the snowballs need to hit the same point, their horizontal and vertical displacements will be equal.
Horizontal displacement for both snowballs:
Sx1 = Vx * t
Sx2 = Vx2 * t
Vertical displacement for both snowballs:
Sy1 = Vy * t + 0.5 * (-g) * t^2
Sy2 = Vy2 * t + 0.5 * (-g) * t^2
Since the horizontal and vertical displacements are equal, Sx1 = Sx2 and Sy1 = Sy2.
Setting these equations equal to each other, we can solve for the angle theta2:
Vx * t = Vx2 * t
Vy * t + 0.5 * (-g) * t^2 = Vy2 * t + 0.5 * (-g) * t^2
Dividing these equations by t, we get:
Vx = Vx2
Vy + 0.5 * (-g) * t = Vy2 + 0.5 * (-g) * t
Substituting the expressions for Vx, Vx2, Vy, and Vy2:
V * cos(theta) = V2 * cos(theta2)
V * sin(theta) + 0.5 * (-g) * t = V2 * sin(theta2) + 0.5 * (-g) * t
Now, we can substitute the given values:
V = 17.9 m/s
theta = 50°
g = 9.8 m/s^2
17.9 * cos(50°) = 17.9 * cos(theta2)
17.9 * sin(50°) + 0.5 * (-9.8) * t = 17.9 * sin(theta2) + 0.5 * (-9.8) * t
cos(50°) = cos(theta2) (since the velocities are equal)
sin(50°) = sin(theta2) (since the times are equal)
Taking the inverse sine of both sides of the second equation:
theta2 = 50°
Hence, the angle at which the second snowball should be thrown to hit the same point as the first snowball is also 50°.
Therefore, the answer is 50°.