Question

If Hiawatha held his bow 1.5 m off the ground and shot the arrow at a 45° angle with an initial velocity of 40 m/s, then the arrow’s height (in meters) off the ground can be modeled by the equation h(t)=−9.8t2+28t+1.5.

Answer 1

The arrow hits the top of a 3 meters tall wigwam at 0.054 seconds

How to determine the time to reach 3 meters?

The complete question is added as an attachment

The function is given as:

$h(t) = -9.8t^2 + 28t + 1.5$

Set the height to 3

$-9.8t^2 + 28t + 1.5 = 3$

Subtract 3 from both sides

$-9.8t^2 + 28t - 1.5 = 0$

Apply the following quadratic formula

$t = \frac{-b \pm \sqrt{b^2- 4ac}}{2a}$

So, we have:

$t = \frac{-28 \pm \sqrt{28^2- 4*-9.8*-1.5}}{2*-9.8}$

This gives

$t = \frac{-28 \pm \sqrt{725.2}}{-19.6}$

This gives

$t = \frac{-28 \pm 26.93}{-19.6}$

Expand

$t = \frac{-28 + 26.93}{-19.6}$ and $t = \frac{-28 - 26.93}{-19.6}$

This gives

t = 0.054 and t = 2.80

0.054 is less than 2.80

This means that the arrow hits the top of a 3 meter tall wigwam at 0.054 seconds

Read more about quadratic functions at

brainly.com/question/27958964

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