Question

A projectile is launched straight up from the ground with an initial velocity of 120 ft/s. If acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft?

Answer 1

Answer:

It will take 2.5 s to reach a height of 200ft.

Step-by-step explanation:

Answer 2

Answer:

The projectile will reach a height of 200 ft after 1.910 seconds and 13.090 seconds of being launched.

Step-by-step explanation:

This projectile is experimenting a free fall, that is, an uniform accelerated motion of the projectile due to gravity and in which effects of air friction and Earth's rotation are neglected. Given that initial and final heights, initial velocity and gravitational acceleration are known, we need to calculate time by solving the appropriate equation of motion described below:

$y = y_{o} +v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of projectile, measured in feet.

$y$ - Final height of projectile, measured in feet.

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

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

As we must remember, quadratic functions have two roots and if we get that $y_{o} = 0\,ft$, $y = 200\,ft$, $v_{o} = 120\,\frac{ft}{s}$ and $g = -16\,\frac{ft}{s^{2}}$, then, the quadratic function is:

$-8\cdot t^{2}+120\cdot t -200 = 0$

Roots are found by the Quadratic Formula:

$t_{1,2} = \frac{-120\pm \sqrt{120^{2}-4\cdot (-8)\cdot (-200)}}{2\cdot (-8)}$

Whose solutions are:

$t_{1} \approx 13.090\,s$ and $t_{2} \approx 1.910\,s$.

Both roots are physically reasonable, first root represents the instant when the projectile reaches 200 feet when it is moving downwards, whereas second root represents the projectile moving upwards. Therefore, the projectile will reach a height of 200 ft after 1.910 seconds and 13.090 seconds of being launched.