Question

A ball is thrown up vertically returns to the thrower after 6 s . Find:

1) the velocity with which it was thrown up.
2) the maximum height it reaches
3) it's position after 4 s. (Take g =10m/s

Answer 1

The time of ascent = time of descent t =6/2 = 3 seconds

Let the velocity with which the ball is thrown up be 'u' m.s-1 and let the ball reach a maximum height of 'h' m.

(I) For upward velocity

v=0, t= 3s, g = -9.8 m/s2

By using the formula

v=u+gt

= u-9.8*3

u=29.4 m/s

(II) Maximum Height Reaches

s = ut + 1/2 gt2

 = 29.4 * 4 - 1/2 * 9.8 * 42

= 88.2 -44.1 m

= 44.1 m

(III) The position of the ball at 4s

= h - h1

= 44.1 m -  4.9 m

= 39.2 m

Answer 2

thrown vertically and returns in 6 sec

up time = 3 sec

down time = 3 sec

max height at t = 3 sec

velocity at max height = 0 m/s

v = u + at

u = intitial velocity

a = -g = -10 m/s (going upward)

v = 0

t = 3 sec

0 = u + (-10)3

u = 30 m/s

s = ut + (1/2)at^2

h = 30*3 +(1/2)(-10)3^2

h = 90 -45

h = 45m

t = 4s = 1s after reaching max height = 45 m

with 0 initial velocity

now a =g as going downward

s = (1/2)(10)1^2

s = 5 m

height at t = 4 sec

= 45-5

= 40m