Answer:
27°
2.81 s
Explanation:
Projectiles launched at complementary angles have the same range. So you should throw the second snowball at an angle of 27°.
We can prove this with kinematics.
For the first snowball, find the time to land, then the distance traveled.
Given (in the y direction):
Δy = 0 m
v₀ = 31.5 sin 63° m/s
a = -9.8 m/s
Find: t
Δy = v₀ t + ½ at²
(0 m) = (31.5 sin 63° m/s) t + ½ (-9.8 m/s²) t²
t = 5.73 s
Given (in the x direction):
v₀ = 31.5 cos 63° m/s
a = 0 m/s²
t = 5.73 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (31.5 cos 63° m/s) (5.73 s) + ½ (0 m/s²) (5.73 s)²
Δx = 81.9 m
For the second snowball, find the time as a function of θ.
Given (in the y direction):
Δy = 0 m
v₀ = 31.5 sin θ m/s
a = -9.8 m/s
Find: t
Δy = v₀ t + ½ at²
(0 m) = (31.5 sin θ m/s) t + ½ (-9.8 m/s²) t²
t = (31.5 sin θ / 4.9) s
Now find the angle needed to travel the same horizontal distance as the first one.
Given (in the x direction):
Δx = 81.9 m
v₀ = 31.5 cos θ m/s
a = 0 m/s²
t = (31.5 sin θ / 4.9) s
Δx = v₀ t + ½ at²
(81.9 m) = (31.5 cos θ m/s) (31.5 sin θ / 4.9) s + ½ (0 m/s²) (31.5 sin θ / 4.9 s²)
81.9 = 31.5 cos θ (31.5 sin θ / 4.9)
0.405 = cos θ sin θ
0.405 = ½ sin(2θ)
0.809 = sin(2θ)
2θ = 54°
θ = 27°
Part 2
Using the equation found earlier, find the time it takes for the second ball to land.
t = (31.5 sin θ / 4.9) s
t = (31.5 sin 27° / 4.9) s
t = 2.92 s
Find the difference in time:
Δt = 5.73 s − 2.92 s
Δt = 2.81 s
