Question

A body projected vertically up from the ground crosses a point p after t seconds and again after 3t/2 seconds.maximum height reached by the body above p is

Answer 1

Consider the motion from A to C

V₁ = initial velocity at A

a = acceleration = - g

ΔY = vertical displacement = 0

t = time interval = (3t/2) - t = (0.5) t

Using the equation

ΔY = V₁ t + (0.5) a t²

0 = V₁ (0.5) t + (0.5) (- g) (0.5)² t²

V₁ = ( g) (0.5)² t

consider the motion from A to B

V₂ = final velocity at B = 0 m/s

V₁ = initial velocity at A = (0.5)² (gt)

a = acceleration = - g where g = 9.8

h = height of position B above A

Using the equation

V²₂ = V²₁ + 2 a h

0² = ( (0.5)² (gt))² + 2 (- g) h

h = ( (0.5)² (gt))² /(2g)

h = (0.5)³ g t²