Question

a ball thrown up vertically returns to the thrower after 8 s. its height above the ground after 6 s, will be (take g = 10m/s square)

Answer 1

(a) Consider,

upward gravity = -9.8 m/s2

Total time (to and fro)= 6 s, so, upward journey = 6/2 = 3 s

Initial velocity (u) = ?

Final velocity (v) = 0 m/s

From First equation of motion,

Final velocity = Initial velocity +gt

or, 0 = u + (-9.8 X3)

or, u = 29.4 m/s

So, The velocity at which ball is thrown in the upward direction is 29.4 m/s.

(b) the maximum height (hmax) is reached by a ball

Second equation of motion

s = ut + 1/2 gt2

here s = the maximum height (hmax)

or , hmax =(29.4 X 3) + 1/2 (-9.8)(3)2

hmax = 44.1 m

(c) Position of the ball after 4s is

From Second equation of motion

s = ut + 1/2 gt2

or, s = (29.4X4) + 1/2 X (-9.8)(4)2

or, s = 117.6 - 78.4 = 39.2 m

Answer 2

Total time for journey is 8sec
so Time to reach the top is 4sec

v = u +at
0 = u + (-10)x4
u = 40m/s

using v^2 = u^2 + 2as
1600 = 20S
S = 80m

now vel at top = 0
time to reach top is 4sec
so 6sec means 2sec after reaching top

now S=1/2 at^2
= 20m

distance above ground = 80-20= 60m ans