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 samepoint as the first? Note the starting and ending heights are the same. The acceleration of
gravity is 9.8 m/s.
How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.
To solve this problem, we can use the equations of motion and principles of projectile motion.
1. First, let's find the horizontal and vertical components of the initial velocity for the first snowball.
The initial speed (v₀) is given as 32 m/s, and the angle (θ) is given as 52°.
The horizontal component of the initial velocity (v₀x) can be calculated using the equation:
v₀x = v₀ * cos(θ)
Substituting the given values:
v₀x = 32 * cos(52°)
v₀x ≈ 32 * 0.6157
v₀x ≈ 19.6984 m/s (approx.)
The vertical component of the initial velocity (v₀y) can be calculated using the equation:
v₀y = v₀ * sin(θ)
Substituting the given values:
v₀y = 32 * sin(52°)
v₀y ≈ 32 * 0.7934
v₀y ≈ 25.4672 m/s (approx.)
2. Now, let's find the time of flight for the first snowball.
The time of flight (t) can be calculated using the equation:
t = 2 * v₀y / g
Substituting the given values:
t = 2 * 25.4672 / 9.8
t ≈ 5.2003 s (approx.)
3. Next, let's find the horizontal range (R) covered by the first snowball.
The horizontal range (R) can be calculated using the equation:
R = v₀x * t
Substituting the calculated values:
R ≈ 19.6984 * 5.2003
R ≈ 102.2182 m (approx.)
4. Since the starting and ending heights are the same, we know the final vertical displacement (Δy) for the first snowball should be zero.
The vertical displacement (Δy) can be calculated using the equation:
Δy = v₀y * t + (1/2) * (-g) * t^2
Setting Δy to zero and solving for t:
0 = v₀y * t - (1/2) * g * t^2
0 = 25.4672 * t - (1/2) * 9.8 * t^2
This equation is a quadratic equation, and when solved, will give us two possible values for time (t).
Using the quadratic formula or factoring, we find:
t = 0 s (Time when the snowball is initially thrown)
t ≈ 5.2011 s (approx.) (Time when the snowball reaches the same level as the starting point)
5. Finally, let's find the angle at which the second snowball should be thrown.
Since the second snowball needs to reach the same point at the same time, and the horizontal range for the first snowball is 102.2182 m, the horizontal range for the second snowball should also be 102.2182 m.
We can use the equation for horizontal range (R) to find the angle (θ') at which the second snowball should be thrown:
R = v₀x' * t
Substituting the known values:
102.2182 = v₀x' * 5.2011
Solving for v₀x':
v₀x' = 102.2182 / 5.2011
v₀x' ≈ 19.6376 m/s (approx.)
Now we can find the angle (θ') using the equation for the horizontal component of velocity:
v₀x' = v' * cos(θ')
Substituting the known values:
19.6376 = 32 * cos(θ')
cos(θ') ≈ 19.6376 / 32
cos(θ') ≈ 0.6124
Now, we can find the angle (θ') by taking the inverse cosine (arccos) of the value:
θ' ≈ arccos(0.6124)
θ' ≈ 52.6182° (approx.)
So, to make the second snowball hit the same point as the first, you should throw it at an angle of approximately 52.6182° above the horizontal.
6. Finally, to find the time delay between the throws, we subtract the times when the snowballs were initially thrown:
t_delay = t' - t
Substituting the calculated values:
t_delay ≈ 5.2011 - 0
t_delay ≈ 5.2011 s (approx.)
Therefore, the second snowball should be thrown approximately 5.2011 seconds after the first snowball to hit the same point at the same time.