Question #e1645

We're asked to find

• the height of the tower

• the velocity at which the ball strikes the ground

from the given information.

#" "#

Height of Tower

This will be the initial position #y_0# of the ball, and we're going to use the kinematics equation

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

where

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

• #y_0# is the initial position (what we're trying to find)

• #v_(0y)# is the initial velocity of the ball (given as #19.6# #"m/s"#)

• #t# is the time of the motion (given as #5# #"s"#)

• #g = 9.8# #"m/s"^2#

Plugging in known values:

#0 = y_0 + (19.6color(white)(l)"m/s")(5color(white)(l)"s") - 1/2(9.8color(white)(l)"m/s"^2)(5color(white)(l)"s")^2#

#color(red)(ulbar(|stackrel(" ")(" "y_0 = 24.5color(white)(l)"m"" ")|)#

The #color(red)("height"# of the tower is thus #color(red)(24.5color(white)(l)"meters"#.

#" "#

Final Velocity

To find the final velcoity, 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 (#19.6# #"m/s"#)

• #g = 9.81# #"m/s"^2#

• #t# is the time (#5# #"s"#)

Plugging these in:

#v_y = 19.6color(white)(l)"m/s" - (9.8color(white)(l)"m/s"^2)(5color(white)(l)"s")#

#color(blue)(ulbar(|stackrel(" ")(" "v_y = -29.4color(white)(l)"m/s"" ")|)#

The final #color(blue)("velocity"# of the ball is therefore #color(blue)(29.4color(white)(l)"meters per second"# in the #color(blue)("downward direction"#.