Question

Someone help me please:)

Answer 1

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 ⇒ $80t-16t^{2}=0$

∴$16t(5 - t) = 0$

5 - t = 0⇒ ∴ t = 5 seconds

b. To tind the time that the ball will be over 96 ft over the ground substitute in $s(t)=80t-16t^{2}$

$96=80t-16t^{2}$

$16t^{2}-80t+96=0$

$16(t^{2}-5t+6)=0$

$16(t-3)(t-2)=0$

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.