Question

An arrow is shot vertically upward from a platform 35 ft high at a rate of 166 ft/sec. When will the arrow hit the ground? Ignore the air resistance. Use the formula ℎ = −16 2 + 0 + ℎ0. Round your answer to the nearest tenth.

Answer 1

Answer:

The arrow will hit the ground approximately 10.5 seconds after launch.

Step-by-step explanation:

The arrow experiments a free fall motion, which is a particular form of uniform accelerated motion due to gravity and in which air friction and effects from Earth's rotation can be neglected. Then, the height as a function of time ($h(t)$), measured in feet, is obtained by the following kinematic formula:

$h(t) = h_{o}+v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (1)

Where:

$h_{o}$ - Initial height of the arrow, measured in feet.

$v_{o}$ - Initial velocity of the arrow, measured in feet per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in feet per square second.

If we know that $h(t) = 0\,ft$, $h_{o} = 35\,ft$, $v_{o} = 166\,\frac{ft}{s}$ and $g = -32.174\,\frac{m}{s^{2}}$, then the following polynomial is obtained:

$35+166\cdot t -16.087\cdot t^{2}=0$ (2)

Lastly, we find the roots of the given expression by the Quadratic Formula:

$t_{1}\approx 10.526\,s$ and $t_{2} \approx -0.207\,s$

Time is a positive variable and then we conclude that the only solution that is physically reasonable is approximately 10.526 seconds.

The arrow will hit the ground approximately 10.5 seconds after launch.