Answer:
1.47 s
Explanation:
The equations of uniformly accelerated rectilinear motion of upward (vertical ) are:
y = y₀+ (v₀)*t - (1/2)*g*t² Equation (1)
vf² = v₀² -2gy₁ Equation (2)
vf = v₀ -gt Equation (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀: initial vertical velocity in m/s
vf: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 14.3 m/s
y₀ =0
y₁ = 7.8
g = 9.8 m/s²
Calculation of the time it takes for the ball to hit the ground (t)
when the ball hits the ground y = 0
We replace data in the formula (1)
y = y₀ + (v₀)*t - (1/2)*g*t²
0 = 0 + (14.3)*t - (1/2)* (9.8) *t²
4.9t²= (14.3)*t
We divided both sides of the equation by t
4.9t= (14.3)
t = (14.3)/ 4.9
t = 2.92 s
Calculation of the time it takes for the ball to reach the tree branch (t₁)
We replace data in the formula (2)
vf² = v₀² -2gy₁
vf² = (14.3)² -2(9.8)(7.8 )
vf² =204.49 -152.88
vf² = 51.61
[tex]V_{f} = \sqrt{51.61}[/tex]
vf = 7.18 m/s
We replace data in the formula (3)
vf = v₀ -gt₁
7.18 = 14.3 -(9.8)t₁
(9.8)t₁ = 14.3 - 7.18
(9.8)t₁ = 7.116
t₁ = 7.116 / (9.8)
t₁ = 0.72 s
Additional time that will pass before the ball passes the tree branch on the way back (t₂)
t₂ = t - 2*t₁
t₂ = 2.92 s - 2*(0.72)s
t₂ = 2.92 s - 2*(0.72)s
t₂ = 1.47 s
