Answer:
1) v₀x = 13.76 m/s
2) v₀y = 19.66 m/s
3) ymax = 21.199 m
4) X = 55.1746 m
5) and 6) y = 18.4 m
Explanation:
1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s
2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s
3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m
4) We can use this equation
y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²
where y = y₀ = 1.5 m
then
1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²
⇒ 0.02588 x² - 1.42815 x = 0
Solving this equation we get
x₁ = 0 and x₂ = 55.1746 m
The distance between the two girls is 55.1746 m
5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m
y = ? when x = (xmax/2)
ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))
⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s
then we get α' as follows
α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°
v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s
Now we can apply the equation of the path
y = ymax - ((gx²)/(2v₀²))
⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))
⇒ y = 18.4 m
