A ball is thrown from the top of a building upward at an angle of 53◦ to the horizontal and with an initial speed of 19 m/s. The ball is thrown at a height of 39 m above the ground. How long is the ball ""in flight""? The acceleration due to gravity is 9.8 m/s 2

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valeriagonzalez0213 valeriagonzalez0213 · Answer 1 · 2019-10-24T07:03:10+02:00

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