An arrow is shot vertically upward from a platform 48 ft high at a rate of 233 ft per sec. When will the arrow hit the ground? Use the formula: h = −16t2 + v0t + h0.

Just use the quadratic formula, which you learned in Algebra I:

t =

-233±√(233^2 - 4(-16)(48))
--------------------------------
      2(-16)

t = (233±√57361)/32

or,

t = -0.20 or 14.76

We don't allow negative time, so the height is zero (the arrow is back on the ground) after 14.76 seconds.

Well, well, well. An arrow shooting up in the air huh? I hope there are no birds around, they might get a little too excited.

Alright, let's solve this equation. We have the formula h = -16t^2 + v0t + h0, where

h = height of the arrow,
t = time in seconds,
v0 = initial velocity of the arrow,
h0 = height from which the arrow is shot.

Here, h0 is given as 48 ft, v0 is given as 233 ft per sec, and we need to find the time when the arrow hits the ground.

Now, when the arrow hits the ground, the height (h) will be zero. So, we can set the equation to zero:

0 = -16t^2 + 233t + 48.

Now, we just need to solve this quadratic equation to find our answer. I'm not very good with numbers, so let's use the magical quadratic formula for that:

t = (-b ± √(b^2 - 4ac)) / 2a.

Plugging in the values, we get:

t = ( -233 ± √(233^2 - 4 * -16 * 48)) / (2 * -16).

Now, grab your calculator and let's crunch some numbers. I'll wait here with my juggling balls.

*taps foot impatiently*

Did you get it? The two solutions you should have are t = 3.60 seconds or t = 8.40 seconds.

So, according to my calculations, the arrow will hit the ground either after approximately 3.60 seconds or after approximately 8.40 seconds. I bet the arrow is having the time of its life up there. Enjoy the show!

To find when the arrow will hit the ground, we can use the given formula: h = -16t^2 + v0t + h0. Here, h represents the height of the arrow, t represents the time, v0 represents the initial velocity, and h0 represents the initial height.

In this case, the arrow is shot from a platform 48 ft high, so h0 = 48 ft. The initial velocity is given as 233 ft/s, so v0 = 233 ft/s.

The height of the ground is 0 ft, so we can set h to 0 and solve for t:

0 = -16t^2 + 233t + 48

Now, let's solve this quadratic equation for t. We can either use the quadratic formula or factor the equation if possible.

Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a, b, c are coefficients of the equation.

In this case, a = -16, b = 233, and c = 48.

t = (-233 ± √(233^2 - 4 * -16 * 48)) / (2 * -16)

Simplifying further:

t = (-233 ± √(54289 + 3072)) / -32

t = (-233 ± √(57361)) / -32

Now, we have two possible solutions for t:

t1 = (-233 + √57361) / -32
t2 = (-233 - √57361) / -32

Calculating these values will give us the time at which the arrow hits the ground.

it hits the ground, naturally, when h=0. So, just find t when

-16t^2 + 233t + 48 = 0

idon't understand

how to perform this operation