Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
B. Refraction because objects refract in water
there are a reading of something like s-waves and p-waves they tell if an earthquakes coming.
From the calculation, the normal force is 6161.2 N.
<h3>What is the normal force?</h3>
The normal force is given by the expression;
N - mg = ma
Then;
N = mg + ma
m = 84.4 kg
g = 9.8 m/s^2
a = 63.2 m/s2
Now we have;
N = m(g + a)
N = 84.4 (9.8 + 63.2)
N = 6161.2 N
Learn more about normal force:brainly.com/question/15520313
#SPJ1