a body is thrown vertically upward with initial velocity 196m/sec .how long it take to be back in hands of thrower (9.8m/sec)
Answers
Explanation:
Assuming the(edit: acceleration caused by the) force of gravity has a magnitude of exactly 9.8m/s², and no other forces are present, the thing will hit 0m/s in exactly one second.
Solving the 2nd question would require usage of a kinematic equation. Specifically…
v²=u²+2as
This one.
Here,
v is the final velocity of the object
u is the velocity it started off with
a is its acceleration(duh)
s is the position it traverses in the time frame given.
We will be looking for the value of s. To make life easier for us, I shall rearrange the equation in such a way that s is the only value on one side of it and everything else is on the other.
s=(v²-u²)/2a
Much better.
Inserting the values provided, we get
s=(0–96.04)/-19.6
=4.9m
Note that I used “-” to indicate downwards acceleration.
Anyway, the highest point of the object’s trajectory will be 4.9m above wherever it came from.
You’re welcome. Now excuse me, I shall proceed to smile like an maniac at the fact that I just answered a question related to physics(a rudimentary one however) in ages.
Finally.