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

1 Answer
Jun 25, 2017

"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.