Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
