A ball weighing 5 kg is dropped from the top of a 20 m high building. Find the k.E. Possessed by the ball when it reaches half the way to the ground?
Answers
Answer:
Given: Mass = 5kg,
Initial velocity = 0 m/s (dropped)
Total height = 20 m
To find : KE when ball reaches half height
(i.e 10m)
Calculation : Using kinematic equation:
v^2 - u^2 = 2*g*h
=> v^2 - 0 = 2*10*10
=> v^2 = 200.
Now KE = 1/2 m *(v^2)
=> KE = 1/2 *5*200
=> KE = 500 JOULES.
SO THE ANSWER IS 500 JOULES.
INDEX : g = gravity, h = height , v = final velocity, u = initial velocity, KE = Kinetic energy.
AnswEr :
- Mass ( m ) = 5 Kg
- Height ( h ) = 20 m
- Initial Velocity ( u ) = 0 m/s²
- Kinetic Energy = ?
⋆ Using Third Equation of Motion :
» v² - u² = 2gh
- Plugging the Values
» v² - ( 0 )² = 2 × 10 × 20
» v² - 0 = 200
» v² = 200 m/s²
⋆ We know the Formula for Kinetic Energy :
⇒ K.E. = 1 /2 × Mass × (Final Velocity)²
- Plugging the Values
⇒ K.E. = 1 /2 × 5 × 200
⇒ K.E. = 5 × 100
⇒ K.E. = 500 Joule
჻ Kinetic Energy of 500 Joule is Possessed by the ball when it reaches half the way to the ground.