Question

Ted throws an object into the air with an initial vertical velocity of 54 feet per second from a platform that is 40 feet above the ground. About how long will it take the object to hit the ground?

Answer 1

PERFECT application of the general equation of things
that are moving up and down in gravity:

H(t) = H₀ + v₀t + 1/2 g t²

t = time after the toss
H(t) = height at any time after the toss
H₀ = height from which it's tossed
v₀ = initial speed
g = acceleration of gravity

I'm going to call 'UP' = the positive direction .
So ...

H₀ = +40 ft
v₀ = +54 ft/sec
g = -32 ft/sec²

When the object hits the ground, H(t) = 0 .
So you have ...

0 = 40 + 54 t - 16 t²

Let me re-write that for you, to make it a
little easier and more familiar:

- 16 t² + 54 t + 40 = 0

Divide each side by -2 : 8 t² - 27 t - 20 = 0

Now go to your toolbox, pull out your quadratic formula,
solve that quadratic equation, and you'll get the usual two
solutions.

The positive one is the one you want.

The negative one refers to time before the toss, and may be discarded.

Answer 2

The height of the object as a function of time is given by:

`h(t) = -16t^2+vt+h` .....[1]

where,

h(t) is the height of the object t second after it is thrown.

v is the initial velocity

h is the initial height.

As per the statement:

Ted throws an object into the air with an initial vertical velocity of 54 feet per second from a platform that is 40 feet above the ground

⇒v = 54 ft/sec and h = 40 feet above the ground.

Substitute in [1] we have;

`h(t) = -16t^2+54t+40`

we have to find how long will it take the object to hit the ground.

Substitute h(t) = 0

then;

`-16t^2+54t+40 = 0`

⇒
`-8t^2+27t+20=0`

Factorize this equation:

`-8t^2+32t-5t+20=0`

⇒
`-8t(t-4)-5(t-4)=0`

⇒
`(t-4)(-8t-5)=0`

By zero product property we have;

t-4 = 0 and -8t-5 = 0

⇒t = 4 and -8t = 5

⇒ t= 4 and t = - 0.625

Since, time cannot be in negative;

⇒ t= 4 sec

therefore, 4 seconds will it take the object to hit the ground