Answer:
81.1 m
Explanation:
Given:
x₀ = 0 m
y₀ = y = 0 m
v₀ₓ = 29.5 cos 33.0° m/s
v₀ᵧ = 29.5 sin 33.0° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
Find:
x
First, find the time it takes to land.
y = y₀ + v₀ᵧ t + ½ aᵧt²
0 = 0 + (29.5 sin 33.0) t - 4.9 t²
t = (29.5 sin 33.0) / 4.9
t = 3.28 s
Now in the x direction:
x = x₀ + v₀ₓ t + ½ aₓt²
x = 0 + (29.5 cos 33.0) (3.28) + 0
x = 81.1 m
The outfielder was 81.1 meters away.
Answer:
vi = 2.83 √gR
Explanation:
For this exercise we can use the law of conservation of energy
Let's take a reference system that is at point A, the lowest
Starting point. Lower, point A
Em₀ = Ki = ½ m vi²
Final point. Higher, point B
= K + U
It indicates that at this point the kinetic energy is ki / 2 and the potential energy is ki / 2
K = ki / 2
U = m g (2R)
Energy is conserved so
Em₀ = Em_{f}
½ m vi² = ½ (1/2 m vi²) + m g 2R
½ m vi² (1- ½) = m g 2R
vi² = 4 g 2 R
vi = √ 8gR = 2 √2gR
vi = 2.83 √gR
The amplitude of a wave can be obtained by measuring the distance from the resting position of the wave to its crest. The resting position is half of the distance from the crest to the trough. Given that the distance between the crest and the trough is 3 meters, the amplitude should be half of that, which is 1.5 meters.
Answer:
all of the above characteristics
