Question

A waterfall has a hit of 1500 ft. A pebble is thrown upward at the velocity of 24ft per second. The hot, h, of the pebble after t seconds is given by the equation h=-16t+24t+1500. How long after the pebble is thrown will it hit the ground.

Answer 1

Answer:

A waterfall has a height of 1500 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 24 feet per second. The​ height, h, of the pebble after t seconds is given by the equation h =−16t^2+24t+1500. How long after the pebble is thrown will it hit the​ ground?

Pebble will hit the ground after 10.46 seconds.

Step-by-step explanation:

Given:

The​ height, "h" of the pebble after t seconds, "h" = −16t^2+24t+1500

We have to find the time it will take to hit the ground.

For this we have to put h = 0 and solve the quadratic.

Quadratic formula:

⇒ Standard equation : $ax^2+bx+c=0$

⇒ $x =\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}$

Now,

Solving the above equation with quadratic formula after comparing its values with the standard equation.

⇒ $a=-16,\ b=24,\ c=1500$

⇒ $t =\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}$

⇒ $t =\frac{-24\pm \sqrt{(24)^2-4(-16)(1500)} }{2(-16)}$

⇒ $t =\frac{-24\pm \sqrt{576+96,000} }{-32}$

⇒ $t =\frac{-24\pm(310.76)}{-32}$

⇒ $t =\frac{-24+(310.76)}{-32}$                         ⇒ $t =\frac{-24-(310.76)}{-32}$

⇒ $t =-\frac{286.76}{32}$                                ⇒ $t =\frac{334.76}{32}$

⇒ $t=-8.96\ sec$                            ⇒ $t=10.46\ sec$

Discarding the negative values.

The pebble will hit the ground after 10.46 seconds.