Respuesta :
Answer:
Vy = 0
y(max) = 137,76 m
x(max) = 318 m
Explanation:
We are dealing with a projectile movement
And from problem statement we know:
θ = 60⁰ then sin θ = sin 60⁰ = √3/2 and cos 60⁰ = 1/2 and g = 9.8 m/sec²
V₀ = 60 m/sec
V₀x = V₀ *cosθ Vx = V₀x Vx = Constant then Vx = V₀ *cosθ
Then Vx = 60* 1/2 Vx = 30 m/sec
V₀y = V₀ *sinθ ⇒ V₀y =60*√3/2 ⇒ V₀y = = 30*√3 m/sec
Vy = V₀y * sin θ - g*t
When Vy = 0 (maximum height point) we are half of the way for the ball to hit the ground, then
Vy = 0 ⇒ V₀y - g*t = 0 ⇒ 30*√3 (m /sec) = 9,8 (m/sec²)* t
t = 30*√3/9.8 t = 5.30 sec
y(max) = y₀ + V₀y*t - 1/2 * g*t²
By substitution:
y (max) = 0 + 30*√3 * 5.30 - 0,5* 9.8* (5.3)²
y(max) = 275,40 - 137,64
y(max) = 137,76 m
And finally x(max)
x(max) = Vx *t = 30* 2*5,3
x(max) = 318 m
