Answer:
a. The time to strike the ground is 5 seconds
b. The time for the ball to be over than 96 ft is 1 second
Step-by-step explanation:
a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero
∴ s(t) = 0 ⇒ [tex]80t-16t^{2}=0[/tex]
∴[tex]16t(5 - t) = 0[/tex]
5 - t = 0⇒ ∴ t = 5 seconds
b. To tind the time that the ball will be over 96 ft over the ground substitute in [tex]s(t)=80t-16t^{2}[/tex]
[tex]96=80t-16t^{2}[/tex]
[tex]16t^{2}-80t+96=0[/tex]
[tex]16(t^{2}-5t+6)=0[/tex]
[tex]16(t-3)(t-2)=0[/tex]
t = 3 sec. and t = 2 sec.
The ball will be at height 96 ft at time 2 sec. and 3 sec.
So the ball will be at height over than 96 = 3 - 2 = 1 sec.
