Aball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. if it is dropped from this height, what will be the kinetic energy of the ball
when it reaches 1.5 meters above the ground?
a.1.2 joules
b.2.2 joules
c.3.4 joules
d.3.6 joules
e.4.3 joules

3.4 Joules

Explanation:

the KE= 1/2 mv^2

By the equation v^2  = u^2  + 2as

where v is the final velocity

u is the initial velocity=0

a is the acceleration=-9.81

s is the displacement = 3.25-1.5=1.75(downwards)=-1.75

v^2= 2(-9.81)(-1.75)

v^2=34.335

KE: 1/2(0.2)(34.335)

=3.4335

=3.4 Joules( rounded to 1 decimal places)

Kinetic energy, KE = 3.43 Joules

Explanation:

It is given that,

Mass of the ball, m = 0.2 kg

Initially, it is placed at a height of 2.25 meters. We need to find the kinetic energy of the ball  when it reaches 1.5 meters above the ground. It can be calculated using the conservation of energy.

To find,

Kinetic energy of the ball when it reaches 1.5 meters above the ground

KE = 3.43 Joules

So, the kinetic energy of the ball is 3.43 joules.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6