An object is thrown upwards with a velocity of 19.6m/s from the ground. How much time will it take to return to its original position?
Answers
Explanation:
An object is thrown vertically upwards with a velocity of 19.6 m/s. What is the displacement of, and distance travelled by, the object after 3 seconds?
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We know ,s= ut - 1/2at^2 (as the body is projected upwards)
S = 19.6* 3 + 1/2 * 9.8 * 3^2
=14.7 mt (which is displacement)
Now we will find at what theq body reach its maximum height i.e., V =0 ,, V= U - gt ( as the body is moving against gravity so negative sign has been used)
0 = 19.6 - 9.8 t
t = 2 sec
Now maximum height attained s= ut - 1/2 gt^2
= 19.6* 2 - 1/2 * 9.8 * 2^2
=19.6 mt
Now while coming down ,distance travelled in (3–2) sec
S= ut + 1/2 gt^2
=0 * 1 + 1/2 * 9.8 *1^2
=4.9 mt
Therefore total distance travelled by the body is ( 19.6+ 4.9) = 24.5 mt
Final ans : displacement = 14.7 mt and distance = 24.5 mt.
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A body is vertically thrown upward with the initial velocity of 19.6 m/sec. What height will it reach?
19.6 = 2*9.8 and we know g = 9.8 ms–2
Therefore in 2 second the speed will be zero .
Distance moved up in 2 second is average speed * time
= (19.6/2)*2 = 19.6 m.
In the next second, its speed increases from 0 to 9.8 m/s .( moves down)
Distance travelled in one second is average speed * time =(9.8/2)*1 = 4.9 m/s
Total disstance =( 19.6+ 4.9 ) = 24.5 m .
Displacement = (19.6 - 4.9) = 14.7
Answer:
Explanation:
Maximum height : 2.5 Marks
Total time : 2.5 Marks
(i) Given
u
=
19.6
m
/
s
,
g
=
–
9.8
m
/
s
2
,
v
=
0
U
s
i
n
g
e
q
u
a
t
i
o
n
v
2
=
u
2
+
2
a
s
,
w
e
g
e
t
0
=
19.6
2
+
(
2
×
(
–
9.8
)
×
h
)
h
=
19.6
m
(ii) Time taken to reach the maximum height can be calculated by the equation,
v
=
u
+
g
t
0
=
19.6
+
(
–
9.8
×
t
)
t = 2 s
In the same time, it will come back to its original position.
Total time =
2
×
2
=
4
s