Question

A ball is thrown straight up, from 3 m above the ground, with a velocity

of 14 m/s. The equation to model this path is h(t)= -5t^2 + 14t + 3. How
would you find when the ball is 8 m above the ground?
Your answer
O This is a required question
If you can, find the solution to the above problem and briefly describe
how you found your solution.
Your answer

Answer 1

The solution of the problem is

Given that:

The equation is
`h(t)=-5t^2+14t+3` , where
`h(t)` is height .

The ball is
`8m` above the ground so
`h=8m` .

Now,

Substitute the value of height in given equation,

`h=-5t^2+14t+3\\\\8=-5t^2+14t+3`

Subtract
`8` on both side to obtain the quadratic equation,

`-5t^2+14t+3-8=8-8\\\\-5t^2+14t-5=0`

Multiply minus sign in both sides,

`5t^2-14t+5=0`

Solve the quadratic equation ,

Where,

`a=5,b=-14,c=5`

`x=-b +\frac{\sqrt{b^(2)-4ac } }{2a} \\\\ x=-b -\frac{\sqrt{b^(2)-4ac } }{2a}`

Substitute the known values in the formula,

`x=(14+√((-14)^2-4(5)(5)) )/(2(5)) \\x=(14+√(196-100) )/(10) \\\\x=(14+√(96) )/(10) \\\\x=(14+√(2*2*2*2*2*3) )/(10) \\\\x=(14+(4√(6)) )/(10) \\\\x=(7+2√(6) )/(5)`

Similarly,

`x=(7-2√(6) )/(5)`