Question

A ball is thrown vertically upward with speed 65m/s . the distance covered by the in 7th sec is?

Answer 1

`"Total distance" = 216` `"m"`

Explanation:

We're asked to find the total distance traveled of a projectile after `7` `"s"` with an initial velocity of `65` `"m/s"` upward.

We can do this by adding together two distances:

1. The distance from ground level (or from wherever it was thrown) to its maximum height.

2. The distance from its maximum height to its height at `t = 7` `"s"` (on the way back down)

(1)

To find its maximum height, we can use the equation

`(v_y)^2 = (v_(0y))^2 + 2a_y(Deltay)`

where

• `v_y` is the velocity at its maximum height, which is `0`

• `v_(0y)` is the initial velocity (`65` `"m/s"`)

• `a_y` is the constant acceleration, which is `-g`, equal to `-9.8` `"m/s"^2`

• `Deltay` is its height (in `"m"`)

Plugging in known values, we have

`0^2 = (65color(white)(l)"m/s")^2 + 2(-9.8color(white)(l)"m/s"^2)(Deltay)`

`(19.6color(white)(l)"m/s"^2)(Deltay) = 4225color(white)(l)"m"^2"/s"^2`

`Deltay = color(red)(215.6` `color(red)("m"`

It thus traveled a distance `215.6` `"m"` upward.

(2)

Now, let's find its position at `t = 7` `"s"`, and subtract this from its maximum height to find how far it went downward.

To do this, we can use the equation

`Deltay = v_(0y)t - 1/2g t^2`

Plugging in values, we have

`Deltay = (65color(white)(l)"m/s")(7color(white)(l)"s") - 1/2(9.8color(white)(l)"m/s"^2)(7color(white)(l)"s")^2`

`Deltay = 214.9` `"m"`

Thus, it traveled a downward distance of

`215.6color(white)(l)"m" - 214.9color(white)(l)"m" = color(green)(0.661` `color(green)("m"`

Total distance

The total distance it travels is the distance traveled upward plus the distance downward:

`"Total distance" = color(red)(215.6` `color(red)("m")` `+ color(green)(0.661` `color(green)("m"` `= color(blue)(216` `color(blue)("m"`

I'll leave it to no decimal places, even though it's practically the same as the upward distance.