**Answer:**

t= 4.765 s : <em>**Flight time**</em>

**Explanation:**

**Known data**

v₀ = 19 m/s, Initial speed

α₀= 53° , Initial angle with the horizontal

y₀ = 39 m , Initial height of the ball above the ground

g = 9.8 m/s² : acceleration due to gravity

**Initial speed components**

v₀x = v₀cosα₀ = 19*cos53° = 11.43 m/s : v₀ x-component

v₀y = v₀sinα₀ = 19*sin53° = 15.17 m/s : v₀ y-component

**Kinematic equations of the parabolic movement**

**x: Uniform movement**

x= v₀x *t :General Equation of the horizontal movement

x= (11.43 )*t Equation (1) of the horizontal movement of the ball

**y: Uniformly accelerated movemen**t

y= -(1/2) gt² + (v₀y)t + y₀ :General Equation of the vertical movement

**y= -4.9t²+ (15.17) t + 39 :Equation (2) of the vertical movement of the ball **

<h3><em>

**Flight time calculation**</em></h3>

We calculate how long the ball lasts in the air by making y = 0 in equation (2)

0= -4.9t²+ (15.17) t + 39 We multiply the equation by (**-1**)

4.9t²- (15.17) t - 39 = 0 **quadratic equation**

Solving the quadratic equation we get :

t₁ = 4.765

t₂ = -1.67

Since the time can only be positive, then the flight time is equal to 4.765 seconds

t= 4.765s <em>**Flight time**</em>