Given data:
g = 10m/s2
Total time T = 6sec
Ta = Td = 3sec
(a) Final velocity at maximum height v = 0
From first equation of motion:-
v = u - gta
u = v + gta
= 0 + 10 x 3
= 30m/s
The velocity with which the stone was thrown up is 30m/s.
(b) From second equation of motion
\( s = ut_a - \frac{1}{2}g(t_a)^2\)
\( = 30 \times 3 - (1/2) \times 10 \times (3)^2\)
\( = 90-45 = 45m\)
The maximum height stone reaches is 45 m.
(c) In 3sec, it reaches the maximum height.
Distance travelled in another 1sec = s’
\( s = ut_a - \frac{1}{2}g(t_a)^2\)
\( s = 0 + 10 \times 1 \times 1 \)
s = 5m.
The distance travelled in another 1sec = 5m.
Therefore in 4sec, the position of point p (45 – 5)
= 40m from the ground.
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