Question #f7f80

We're asked to find

• (A) the initial speed of the ball

• (B) the maximum height of the ball

given its time of flight (#6.00# #"s"#).

#" "#

• (A) Initial speed

To find the necessary initial speed, we recognize that the particle will be at its maximum height at #t = 3.00# #"s"# (halfway split time between upward motion and downward motion), and at this point, the instantaneous #y#-velocity is #0#.

We can then use the equation

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

where

• #v_y# is the #y#-velocity at time #t# (#0#, maximum height)

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

• #g = 9.81# #"m/s"#

• #t# is the time (#3.00# #"s"#)

Plugging in known values:

#0 = v_(0y) - (9.81color(white)(l)"m/s"^2)(3.00color(white)(l)"s")#

#color(red)(ulbar(|stackrel(" ")(" "v_(0y) = 29.4color(white)(l)"m/s"" ")|)#

#" "#

• (B) Maximum height reached

To find this, we can use the kinematics equation

#Deltay = ((v_y + v_(0y))/2)t#

where

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

• #v_y# still equals #0# (maximum height)

• #v_(0y) = color(red)(29.4color(white)(l)"m/s"#

• #t = 3.00# #"s"#

Plugging in known values:

#Deltay = ((0+color(red)(29.4color(white)(l)"m/s"))/2)(3.00color(white)(l)"s")#

#color(blue)(ulbar(|stackrel(" ")(" "Deltay = 44.1color(white)(l)"m"" ")|)#