Answer:
t= 4.765 s : Flight time
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 movement
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
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 Flight time
