3) A body moves with a uniform acceleration of 2 m/s2. If the initial velocity of the
body is 10 m/s, find the velocity of the body after 2 seconds. (Ans. 14 m/s)
(4) A stone is released from the top of a tower of height 78.4 m. Find the time taken
by the stone to reach the ground. (g = 9.8 m/a“)
(Ans. 4 s)
(5) A stone is released from the top of a tower of height 200 m. Find its velocity arter
4 seconds. (g = 9.8 m/s)
(Ans. - 39.2 m/s)
(6) A car starting from rest with a uniform acceleration covers a distance of min
10 seconds from the start. Find its (i) acceleration and (ii) velocity after 10 seconds
from the start.
(Ans, (i) 1 m/s2 (ii) 10m/s)
(7) A car initially at rest to given a uniform acceleration of 0.5 m/s' for 20 seconds.
The car then moves with a uniforın velocity for 20 seconds. The brakes are then
applied and the car is brought to rest in 10 seconds. Find the total distance covered
by the car.
(Ans. 350 m)
(8) The velocity of a car decreases from 20 m/s to 10 m/s in a certain time interval.
Find the retardation of the car if the distance covered by the car during this time
is 300 m.
(Ans. 0.5 m/s)
(9) A stone is released from the top of a tower of height 200 m. Find the distance
covered by the stone during the fifth second of its motion (g = 9.8 m/s)
(Ans. 44.1 m)
Answers
Answered by
2
For I
st
stone u
h
=10 m/s
u
v
=0;s
v
=−78.4 m;g=−9.8 m/s
2
s
v
=u
v
t+
2
1
at
2
⇒−78.4=
2
−9.8
×t
2
t
2
=16 sec.⇒t=4 sec.
Distance travelled in horizontal direction in 4 s
s
n
=10×4=40 m
For II
nd
stone →u
h
=20 m/s;u
v
=0;s
v
=−78.4 m;g=−9.8 m/s
2
It will also take same time as the vertical distance travelled is same. Distance travelled in horizontal direction by II
nd
stone in 4 s
s
n
=20×48=80 m
total distance between stones =40+80=120 m
st
stone u
h
=10 m/s
u
v
=0;s
v
=−78.4 m;g=−9.8 m/s
2
s
v
=u
v
t+
2
1
at
2
⇒−78.4=
2
−9.8
×t
2
t
2
=16 sec.⇒t=4 sec.
Distance travelled in horizontal direction in 4 s
s
n
=10×4=40 m
For II
nd
stone →u
h
=20 m/s;u
v
=0;s
v
=−78.4 m;g=−9.8 m/s
2
It will also take same time as the vertical distance travelled is same. Distance travelled in horizontal direction by II
nd
stone in 4 s
s
n
=20×48=80 m
total distance between stones =40+80=120 m
Answered by
2
Answer:
3)14 m/s
4)4s
5)39.2 m/s
You already have the rest of the answers bruh :P
Explanation:
3)u=10
a=2
t=2
v=u+at=10+4=14m/s
4)g=9.8
u=0
s=78.4
s=1/2gt^2
t=sqrt(2s/g)=√2*78.4/9.8=4 s
5)s=200 m
u=0
t=4 s
g=9.8m/s^2
v=gt=9.8*4=39.2 m/s
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