Question #04fd1

We use the kinematic equation connecting various quantities of interest. The maximum height is reached when upwards velocity is #=0#
1. To calculate Max height #h# the kinematic equation is
#v^2-u^2=2ah#
Final velocity #v# after time #t#, #a# is acceleration and initial velocity is #u#

Inserting given values we get, remember, direction of acceleration due to gravity is opposite to the direction of initial velocity

#0^2-30^2=2xx(-10)h#
#=>20h=900#
#=>h=900/20=45m#
2. To calculate time taken to reach maximum height we use the other kinematic equation
#v=u+at#

Inserting given values we get

#0=30-10xxt#
#=>10xxt=30#
#=>t=30/10=3s#

The initial Kinetic energy (where potential energy is zero) becomes only potential energy (where the ball reaches the max height and its speed is zero).
#1/2 m v^2 = mgh #
but, since the mass is the same
#1/2 v^2 = gh#
and
# h= v^2/(2g)#
# h = (900 m^2/s^2)/(2 10 m/s^2)= 45 m#
from the second law of motion
#v= v°-at#
#0= 30 m/s-10m/s^2 t#
#t= 3 s#
now let's verify in the first law of motion
#s= v°t -1/2 g t^2#
#45m = 30 m/s 3s - 1/2 10 m/s^2 (3s)^2#
45 = 45 m