Question

a particle is projected horizontally from the top of a tower of height H with kinetic energy K. The particle lands on the ground with kinetic energy 2K. during its flight,the kinetic energy of the particle as a function of its height h above the ground is given by:

a. (1+h/H)K
b. (2-h/H)K
c. (1+(h/H)²)K
d. (2-(h/H)²)K​

Answer 1

Answer:

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Answer 2

Given,

particle projected upwards

the initial kinetic energy of the particle = K

the kinetic energy of the particle when it lands on the ground = 2K

To find,

The particle's kinetic energy is a function of height, h above the ground.

Solution,

When the particle is thrown initially,

K = mgH = 1/2 m$v^2$ (where v is the particle's velocity)

⇒ mg = K/H _____ (1)

Now, let K$_f$ be the kinetic energy of the particle above the ground at a height h.

Now since we know at any given point,

ΔU + ΔK = 0 (law of conservation of energy)

⇒ mg (h-H) + K$_f$ - K = 0

⇒K$_f$ = K - mg (h-H)

⇒K$_f$ = K + mg (H-h) _____(2)

from equation (1) and (2) we get,

K$_f$ = K + K/H (H-h)

    = K ( 1 + $\frac{H-h}{H}$)

    = K (1 + 1 -h/H)

⇒ K$_f$ = K ( 2-h/H)     (option b)