You throw a softball (of mass 300 g) straight up into the air. it reaches a maximum altitude of 10.1 m and then returns to you. what is the gravitational potential energy of the softball at its highest position? assume the ball departed from and returned to ground level.
assume no energy is lost by the softball while it is in the air. what is the kinetic energy of the softball as soon as it leaves your hand? answer in units of j.
what is the kinetic energy of the softball when it returns to your hand? answer in units of j.
what is the speed of the ball? answer in units of m per s.

Answer

given,

mass of the ball = 300 g

                           = 0.3 Kg

maximum altitude of the ball = 10.1 m

Potential energy of the softball:

PE = m g h

PE = 0.3 x 9.8 x 10.1

PE = 29.69 J

Maximum kinetic energy at which the ball is released from hand to reach height of 10.1 m is 29.69 J.

Kinetic energy at which ball returns to your hand is equal to 29.69 J.

speed of the ball

       v² = 197.93

      v = 14.069 m/s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.