Question

5)b. A body of mass 5 kg moves vertically up with a velocity of 10 m/s from the ground. Calculate its kinetic energy and the maximum height attained.

Answer 1

KE = 1/2 mv^2

KE = 1/2 ( 5×10^2)

KE = 1/2 (500)

KE = 250 Jules Ans.

For the projectile motion, the maximum height attained by an object is given as:

h_max = (v^2 sin^2theta)/(2g)

where, hmax is the maximum height attained during the projectile motion

v: initial velocity = 10 m/sec

g: acceleration due to gravity (9.8 m/sec^2), and

theta : 90 degrees in this case (as the body is 'moves up')

and the time taken to reach maximum height is given as :

h_max = (v^2 sin theta)/2g

{theta= 90°= π/2}

thus,

h_max = [(10)^2 sin^2 (π/2)]/(2x9.8) = 18.292 meter Ans.

Answer 2

A boys of mass 5kg moves vertically up with a velocity of 10m/s from the ground .
so, mass of body , m = 5kg
initial speed of body , u = 10m/s
so, initial kinetic energy , K.E = 1/2 mu²
= 1/2 × 5 × 10² = 250 J

at maximum height , body becomes rest
so, final speed of body, v = 0
so, final kinetic energy = 0

let height attained by body is h
use formula, v² = u² + 2as
here, S = h, u = 10, v= 0 and a = -g = -10m/s²
so, 0 = 10² - 2 × 20 × s
s = 5 m