Question

A bird sees a worm on the ground 50 m. below. The bird dives for the worm at a speed of 6 m/sec. The function h\left( t \right) = - 4.9{t^2} + 6t + 50h ( t ) = − 4.9 t 2 + 6 t + 50, where h(t) is the height of the bird at time t.
How long does the worm have before he is eaten? Round to the nearest tenth of a second.

Group of answer choices

1.9 seconds

3.9 seconds

12 seconds

4.9 seconds

Answer 1

Given:

The height of the bird at time t is given by the function $h(t)=-4.9t^2+6t+50$

We need to determine the time it takes the worm to be eaten by the bird.

Time taken:

The time can be determined by substituting h(t) = 0 in the function.

Thus, we have;

$0=-4.9t^2+6t+50$

Switch sides, we get;

$-4.9t^2+6t+50=0$

Let us solve the equation using the quadratic formula.

Thus, we get;

$t=\frac{-6 \pm \sqrt{6^{2}-4(-4.9) 50}}{2(-4.9)}$

Simplifying, we get;

$t=\frac{-6 \pm \sqrt{36+980}}{-9.8}$

$t=\frac{-6 \pm \sqrt{1016}}{-9.8}$

$t=\frac{-6 \pm 31.87}{-9.8}$

The values of t are given by

$t=\frac{-6 + 31.87}{-9.8}$   and   $t=\frac{-6 - 31.87}{-9.8}$

$t=\frac{25.87}{-9.8}$    and   $t=\frac{-37.87}{-9.8}$

$t=-2.4$    and   $t=3.9$

Since, the value of t cannot be negative, then $t=3.9$

Thus, the time taken by the bird to eat the worm is $t=3.9$ seconds.

Hence, Option B is the correct answer.