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

Answer:

Maximum height of the object is 94 feet.

Maximum time taken is 2.9 seconds.

Step-by-step explanation:

We have,

Height of the bridge is $h=-16t^{2}+16t+90$.

i.e. $h=-16t^{2}+16t+90$

i.e. $\frac{dh}{dt}=-32t+16$

Equating $\frac{dh}{dt}=0$ gives,

$-32t+16=0$

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

i.e. $t=\frac{-16}{-32}$

i.e. $t=\frac{1}{2}$

Again differentiating, we get $\frac{d^2h}{dt^2}=-32<0$.

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

The maximum height of the object will be at the time, $t=\frac{1}{2}$ second.

Thus, the maximum height is given by,

$h=-16\times 0.5^{2}+16\times 0.5+90$

i.e. $h=-16\times 0.25+16\times 0.5+90$

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.