Can any wan help pls
2 answers:
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(a) The stone moves by uniform accelerated motion, with constant acceleration 
directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
from which we find the time t after which the stone reaches the ground:
(c) The velocity of the stone at time t can be written as
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
(d) if the stone has an initial velocity of
, then its law of motion would be
and we can find the time it needs to reach the ground by requiring again y(t)=0:
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
from which we find the time t after which the stone reaches the ground:
(c) The velocity of the stone at time t can be written as
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
(d) if the stone has an initial velocity of
and we can find the time it needs to reach the ground by requiring again y(t)=0:
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.