Question #2027f

We're asked to find

• (i) the maximum height obtained by the mass

• (ii) the time it takes the mass to travel #3.0# #"m"# horizontally

• (iii) the time(s) when the height of the mass is #4.0# #"m"#

(i)

Finding the maximum height:

When the particle is at its maximum height, the instantaneous #y#-velocity is #0#, and we can use the kinematics equation

#ul((v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)#

where

• #v_y# is the #y#-velocity at height #Deltay# (which is #0#)

• #v_0# is the initial speed (given as #25# #"m/s"#)

• #alpha_0# is the launch angle (given as #30^"o"#)

• #g = 9.81# #"m/s"^2#

• #Deltay# is the change in height (what we're trying to find)

Plugging in known values:

#0 = [(25color(white)(l)"m/s")sin(30^"o")]^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#color(red)(ulbar(|stackrel(" ")(" "Deltay = 7.96color(white)(l)"m"" ")|)#

The maximum height is thus #color(red)(7.96color(white)(l)"meters"#.

#" "#

(ii)

The time it takes to travel #3.0# #"m"# horizontally:

To do this, we can use the equation

#ul(Deltax = v_0cosalpha_0t#

where

• #Deltax# is the change in horizontal position (#3.0# #"m"#)

• #v_0 = 25# #"m/s"#

• #alpha_0 = 30^"o"#

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

Plugging in known values:

#3.0color(white)(l)"m" = (25color(white)(l)"m/s")cos(30^"o")t#

#color(red)(ulbar(|stackrel(" ")(" "t = 0.24color(white)(l)"s"" ")|)#

It thus takes #color(red)(0.24color(white)(l)"seconds"# to travel #3.0# #"m"# horizontally.

#" "#

(iii)

The time(s) when the height is #4.0# #"m"#:

We can now use the equation

#ul(Deltay = v_0sinalpha_0t - 1/2g t^2#

where

• #Deltay# is the change in vertical position (#4.0# #"m"#)

• #v_0 = 25# #"m/s"#

• #alpha_0 = 30^"o"#

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

• #g = 9.81# #"m/s"^2#

Plugging in known values:

#4.0color(white)(l)"m" = (25color(white)(l)"m/s")sin(30^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2#

Using the quadratic formula yields

#color(red)(ulbar(|stackrel(" ")(" "t = 0.375color(white)(l)"s"color(white)(l)"and"color(white)(l)t = 2.17color(white)(l)"s"" ")|)#

Therefore, the mass has a height of #4.0# #"m"# when the time is #color(red)(0.375color(white)(l)"seconds"# and #color(red)(2.17color(white)(l)"seconds"#.