Answer:
t₁ = 3 s
Explanation:
In this exercise, the vertical displacement equation is not given
y = 240 t + 16 t²
Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration
Let's replace
864 = 240 t + 16 t²
Let's solve the second degree equation
16 t² + 240 t - 864 = 0
Let's divide by 16
t² + 15 t - 54 = 0
The solution of this equation is
t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1
t = [-15 ±√(225 +216)] / 2
t = [-15 + - 21] / 2
We have two solutions.
t₁ = [-15 +21] / 2
t₁ = 3 s
t₂ = -18 s
Since time cannot have negative values, the correct t₁ = 3s
