Question

Ted throws an object straight up into the air with an initial velocity of 54 ft/s from a platform that is 40 ft above the ground. Use the formula h(t)=−16t2+v0t+h0, where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

Answer 1

Answer:

The time it will take for the object to hit the ground will be 4.

Explanation:

You have:

h(t)=−16t²+v0*t+h0

Being v0 the initial velocity (54 ft/s) and  h0 the initial height (40 ft) and replacing you get:

h(t)=−16t²+54*t+40

To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:

0=−16t²+54*t+40

Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values ​​of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:

$\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}$

In this case you have that:

• a=-16
• b= 54
• c= 40

Replacing in the expression of the calculation of roots you get:

$\frac{-54+\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}$  Expresion (A)

and

$\frac{-54-\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}$ Expresion (B)

Solving the Expresion (A):

$\frac{-54+\sqrt{5476 } }{2*(-16)}= \frac{-54+74}{2*(-16)}=\frac{20}{2*(-16)}=\frac{20}{-32}= -\frac{5}{8}$

Solving the Expresion (B):

$\frac{-54-\sqrt{5476 } }{2*(-16)}= \frac{-54-74}{2*(-16)}=\frac{-128}{2*(-16)}=\frac{-128}{-32}= 4$

These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then the time it will take for the object to hit the ground will be 4.