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 1400
h=−16t2+16t+1400. How long after the pebble is thrown will it hit the ground?
Answer
The pebble hits the ground after 9.8675 s
Step-by-step explanation:
Given
waterfall height = 1400 feet
initial velocity = 16 feet per second
The 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 where
a = 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
