Question

Physics help please

Answer 1

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