# An object is thrown vertically from a height of 7 m at  3 m/s. How long will it take for the object to hit the ground?

Jun 20, 2017

$t = 1.54$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ when the object hits the ground, given that it was thrown vertically upward with a speed of $3 \text{m"/"s}$, at a height of $7$ $\text{m}$.

As soon as the object is let go, it is in a state of free-fall, and the kinematics equations can predict its motion, such as its maximum height reached, time of flight, etc.

To find this time, we can use the equation

$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} {a}_{y} {t}^{2}$

For this equation,

• The height $y$ is ground level, which I'll call $0$ $\text{m}$

• The initial ${y}_{0}$ is $7$ $\text{m}$

• The initial $y$-velocity ${v}_{0 y}$ is $3 \text{m"/"s}$

• The time $t$ is when it is at height $y = 0$, which is what we're trying to find

• The $y$-acceleration ${a}_{y}$ is equal to $- g$, which is -9.8"m"/("s"^2)

Plugging in our known values, we have

$0$ $\text{m} = 7$ $\text{m}$ + (3"m"/"s")t + 1/2(-9.8"m"/("s"^2))t^2

$\left(- 4.9 \text{m"/("s"^2))t^2 + (3"m"/"s}\right) t + 7$ $\text{m}$ $= 0$

t = (-3+-sqrt((3)^2 - 4(-4.9)(7)))/(2(-4.9)) = color(red)(1.54 color(red)("s" (positive solution)
Thus, the object will strike the ground after color(red)(1.54 sfcolor(red)("seconds".