Answer:
t₁ = 2,75 sec
Explanation:
The ball will get h maximum when dh/dt ( its vertical component of the speed is equal to 0. At this moment the ball has flown half of the total flight time
Then
h(t) = - 4t * ( 4*t - 11 )
h(t) = - 16*t² + 44*t
Taking derivatives on both sides of the equation we get:
dh/dt = - 32*t + 44
dh/dt = 0 ⇒ -32*t + 44 = 0
t = 44/32 ⇒ t = 1,375 s
So twice this time
2*t = 2 * 1.375
Total time t₁ = 2,75 sec