Respuesta :
Answer:
a) t₁= 3.256 s, y = 17.89 m , b) v _r = -14.9 m / s
Explanation:
We can work this exercise with kinematic relationships
For the ball
y₁ = y₀₁ + v₀₁ t - ½ g t²
For the platform
y₂ = y₀₂ + v₀₂ t
Where the speed of the ball v₀₁ = 20 m / s and the initial height y₀₁ = 10 m, for the speed of the platform v₀₂ = 3 m / s and the initial height y₀₂ = 4 m
At the point where they find the position of the two is the same
y₁ = y₂
y₀₁ + v₀₁ t - ½ g t² = y₀₂ + v₀₎ t
y₀₁ - y₀₂ + t (v₀₁ –v₀₂) - ½ g t² = 0
We substitute the values
10-4 + t (20-3) - ½ 9.8 t² = 0
6 + 17 t - 4.9 t2 = 0
t² - 3,469 t - 1,224 = 0
We solve the second degree equation
t = [3,469 ±√(3,469 2 - 4 1,224)] / 2
t = [3,469 ± 3.0427] / 2
t₁ = 3.2559 s
t₂ = 0.2132 s
We take the longest time, the smallest time is for the ball with negative initial velocity
t₁= 3.256 s
Let's find out how much the platform has risen in this time
y = y₀₂ + v₀₂ t
y = 4 + 3 3.256
y = 17.89 m
b) let's look for the speed of the ball at this time
v = v₀₁ - g t
v = 20 - 9.8 3.2559
v = -11.9 m / s
The relative speed is
v_r = v –vo2
v_r = -11.9 - 3
v _r = -14.9 m / s
