MATH SOLVE

4 months ago

Q:
# Score! U OT pt1 of 10 ( complete)IVY JUICU IU, U UI TUPUses32.6.1Skill Builder5 Question HelpomeA waterfall has a height of 1400 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height, h, of the pebble after tseconds is given by the equation h - 16t" + 16 + 1400. How long after the pebble is thrown will it hit the ground?The pebble will hit the ground aboutseconds after it is thrown.

Accepted Solution

A:

A waterfall has a height of 1400 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height, h, of the pebble after t seconds is given by the equation h equals negative 16 t squared plus 16 t plus 1400h=−16t2+16t+1400. How long after the pebble is thrown will it hit the ground?AnswerThe pebble hits the ground after 9.8675 sStep-by-step explanation:Given waterfall height = 1400 feetinitial velocity = 16 feet per secondThe height, h, of the pebble after t seconds is given by the equation.[tex]h(t) = -16t^{2}+16t+1400[/tex]The pebble hits the ground when [tex]h = 0[/tex][tex]h=-16t^{2}+16t+1400[/tex] ---------------(1)put [tex]h=0[/tex] in equation (1)[tex]0=-16t^{2}+16t+1400[/tex][tex]-16t^{2}+16t+1400=0[/tex]Divide by -4 to simplify this equation[tex]4t^{2}-4t-350=0[/tex]using the Quadratic Formula wherea = 4, b = -4, and c = -350[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}[/tex][tex]t=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(4)(-350) } }{2(4)}[/tex][tex]t=\frac{4\pm\sqrt{16-(-5600) } }{8}[/tex][tex]t=\frac{4\pm\sqrt{16+5600 } }{8}[/tex][tex]t=\frac{4\pm\sqrt{16+5616 } }{8}[/tex]The discriminant [tex]b^{2}-4ac>0[/tex]so, there are two real roots.[tex]t=\frac{4\pm12\sqrt{39 } }{8}[/tex][tex]t=\frac{4}{8}\pm\frac{12\sqrt{39 }}{8}[/tex][tex]t=\frac{1}{2}\pm\frac{3\sqrt{39 }}{2}[/tex]Use the positive square root to get a positive time.[tex]t=9.8675 s[/tex]The pebble hits the ground after 9.8675 second