Question

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?

Answer 1

`t = 1.54` `"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"m"/"s"`, at a height of `7` `"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_(0y)t + 1/2a_yt^2`

For this equation,

• The height `y` is ground level, which I'll call `0` `"m"`

• The initial `y_0` is `7` `"m"`

• The initial `y`-velocity `v_(0y)` is `3"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` `"m" = 7` `"m"` `+ (3"m"/"s")t + 1/2(-9.8"m"/("s"^2))t^2`

`(-4.9"m"/("s"^2))t^2 + (3"m"/"s")t + 7` `"m"` `= 0`

Using the quadratic formula:

`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"`.