Question #c287b

We're asked to find

• (A) the maximum height the stone reaches

• (B) the velocity at which the stone strikes the ground

• (C) the time taken to reach the maximum height

#" "#

• (A) Maximum Height

To find the maximum height obtained by the stone (and all other information we need), we'll first find the initial velocity of the stone, using the equation

#ul(y = y_0 + v_(0y)t - 1/2g t^2#

where

• #y# is the final position (#0#, ground level)

• #y_0# is the initial position (#85# #"m"#)

• #v_(0y)# is the initial velocity (what we're trying to find)

• #t# is the time (#5# #"s"#)

• #g = 9.81# #"m/s"^2#

Plugging in known values:

#0 = 85# #"m"# #+ v_(0y)(5color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2#

#1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 = 85color(white)(l)"m" + v_(0y)(5color(white)(l)"s")#

#color(red)(v_(0y)) = (1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 - 85color(white)(l)"m")/(5color(white)(l)"s") = color(red)(ul(7.53color(white)(l)"m/s"#

At the stone's maximum height, its instantaneous #y#-velocity is zero, so we can use the equation

#ul((v_y)^2 = (v_(0y))^2 - 2g(y - y_0)#

The variables here we know except for the final height #y#, which will be

#0 = (color(red)(7.53color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(y - 85color(white)(l)"m")#

#color(blue)(ulbar(|stackrel(" ")(" "y = 87.9color(white)(l)"m"" ")|)#

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(B) Final Velocity

To find the final velocity, we can use the equation

#ul(v_y = v_(0y) - g t#

where

• #v_y# is the final velocity (what we're trying to find)

• #v_(0y)# is the initial velocity (found to be #color(red)(7.53color(white)(l)"m/s"#)

• #g = 9.81# #"m/s"^2#

• #t = 5# #"s"#

Plugging these in:

#color(blue)(v_y) = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)(5color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "-41.5color(white)(l)"m/s"" ")|)#

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• (C) Time at Maximum Height

To find the time when the stone reaches its maximum height, we can again use the equation

#ul(v_y = v_(0y) - g t#

where

• #v_y = 0# (at maximum height)

• #v_(0y) = color(red)(ul(7.53color(white)(l)"m/s"#

• #g = 9.81# #"m/s"#

• #t# is the time (what we're trying to find)

And so we have

#0 = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)t#

#color(blue)(ulbar(|stackrel(" ")(" "t = 0.767color(white)(l)"s"" ")|)#