a golf boll is released from rest from the top of the Building. calculate the position in meterof the boll after 2 sec.
Answers
Answered by
3
Given:-
- Initial Velocity of the ball = 0m/s
- Acceleration due to Gravity = +10m/s (Free fall).
- Time = 2s.
To Find:-
- The Position of ball after 2 second.
Formulae Used:-
- S = ut + ½ × a × t²
Where,
s = Distance
u = Initial Velocity
a = Acceleration
t = time
Now,
S = ut + ½ × a × t²
S = 0 × 2 + ½ × 10 × 2²
S = 0 + 5 × 4
S = 20m.
Hence, The position of ball was at 20m.
EXTRA INFORMATION.
- When body is thrown at highest position, V = 0
- When body falls freely u = 0
- The value of g is 9.8m/s²
- The value of g at the center of the earth us zero.
Answered by
5
Answer:
the acceleration of the ball will be g. Initial velocity will be 0.
in T sec. body travels h mts.
by applying equations of motion we get
s=ut+(1/2)gT
2
h=(1/2)gT
2
------[1]
in T/3 sec h 1 =(1/2)gT 2 =(1/2)g( 3T ) 2 =(1/2)g( 9T 2 ) -------[2]
from
[1] and [2] we get h 1 = 9h
distance from point of release.
therefore distance from ground is h−
9h = 98h...
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