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.