Respuesta :
Answer:
a) v₀ = 41.68 m / s , b) t = 3,808 s
, c) vₓ = 34.14 m / s , v_{y} = -13.41 m / s
Explanation:
This is a projectile launch problem
a) To find the initial speed, let's work each component independently
X axis
As there is no friction the speed is constant
v₀ₓ = x / t
Y Axis
v_{y}² = ² - 2 g (y-y₀)
v_{y} = v_{oy} - g t
y = y₀ + v_{oy} t - ½ g t²
Speed components can be found with trigonometry
sin 35 = voy / v₀
cos 35 = v₀ₓ / v₀
v₀ₓ = v₀ cos 35
v_{oy} = v₀ sin 35
At the point where the wall is, time is the same for both movements
t = x / v₀ cos 35
(y-y₀) = v₀ sin35 t - ½ g t²
We replace
y -y₀ = v₀ sin 35 x /v₀ cos 35 - ½ g (x /v₀ cos 35)²
y-yo = x tan 35 - ½ g x² / v₀² cos² 35
Let's clear the initial speed
1 / v₀² = (x tan35 - (y-y₀)) 2 cos² 35 / g x²
v₀² = g x² / 2 (x tan35 - (y-y₀)) cos² 35
Let's calculate
v₀² = 9.8 130² / (2 (130 tan 35 - (21 -1 )) cos² 35
v₀ =√ (165620 / 95.3196)
v₀ = 41.68 m / s
b) Let's use the x component of the movement
t = x / v₀ cos 35
t = 130 / (41.68 cos 35)
t = 3,808 s
c) the velocity on the x axis is constant
vₓ = v₀ cos 35
vₓ = 41.68 cos 35
vₓ = 34.14 m / s
The speed on the y axis is
= v₀ sin35 - g t
v_{y} = 41.68 sin 35 - 9.8 3,808
v_{y} = -13.41 m / s
The negative sign indicates that the ball is descending
