Question

The height h in feet of a projectile launched vertically upward from the top of a 96 foot tall bridge is givem by h=90+16t-16t^2 where t is time in seconds. What is the maximum height an how long will it tale the projectile to strike the ground?

Answer 1

Hi!

h(t)= -16t^2+16t+90
h(t) = -16(t^2 - t) + 90
h(t) = -16(t^2 - t + (1/2)^2) + 90 + 4
h(t) = -16(t - 1/2)^2 + 94
vertex at(1/2 , 94) , since a = -16 < 0 , the parabola opens down, which makes the vertex. the global maximum, so:
Max at(1/2 , 94)
object hits the ground at h = 0, then:
-16(t - 1/2)^2 + 94 = 0
(t - 1/2)^2 = 94/16 = 47/8
t - 1/2 = ± √(47/8)
t = 1/2 ± √(47/8)
t = -1.9 , 2.9 => reject t = -1.9
t = 2.9 sec => time it takes for the projectile to hit the ground

Answer 2

Maximum height of the object is 94 feet.

Maximum time taken is 2.9 seconds.

Explanation:

We have,

Height of the bridge is
`h=-16t^(2)+16t+90`.

i.e.
`h=-16t^(2)+16t+90`

i.e.
`(dh)/(dt)=-32t+16`

Equating
`(dh)/(dt)=0` gives,

`-32t+16=0`

i.e.
`-32t=-16`

i.e.
`t=(-16)/(-32)`

i.e.
`t=(1)/(2)`

Again differentiating, we get
`(d^2h)/(dt^2)=-32<0`.

Thus, by the 'First derivative test of the maxima and minima', we get,

The maximum height of the object will be at the time,
`t=(1)/(2)` second.

Thus, the maximum height is given by,

`h=-16* 0.5^(2)+16* 0.5+90`

i.e.
`h=-16* 0.25+16* 0.5+90`

i.e.
`h=-4+8+90`

i.e.
`h=4+90`

i.e. h = 94 feet.

Hence, the maximum height of the object is 94 feet.

Further, the height of the object when it reaches the ground will be 0 feet.

So, we have,

`0=-16t^(2)+16t+90`

i.e.
`16t^(2)-16t-90=0`

i.e.
`(t+1.9)(t-2.9)=0`

i.e. t = -1.9 sec or t = 2.9 sec

Since, time cannot be negative.

We get, the maximum time taken by the object to reach the ground is 2.9 seconds.