A ball is thrown in vertically upward direction. It returns back to the same position in 2s. Then
the maximum height achieved by the ball is (Take g= 9.8 m/s^2)
(A) 9.8 m
(B) 14.7 m
(C) 4.9 m
(D) 19.6 m
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11th
Physics
Motion in a Straight Line
Problems on Equation of Motion
A ball thrown vertically up...
PHYSICS
A ball thrown vertically upward returns to the thrower after 6 s. The ball is 5 m below the highest point at t = 2 s. The time at which the body will be at same position, (take g = 10m/s
2
).
December 27, 2019avatar
Ashi Kanojia
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ANSWER
Total time of flight(t)=6s
So time to reach the maximum height =3s
if u= projected velocity then:
0=u−g
2
t
⇒u=
2
gt
=
2
60
=30 m/s
Applying equation v
2
=u
2
−2gh(for half journey)
0=900−20h
⇒h=45m
When the ball is at height 5m from highest point, it's height from ground =45−5=40m
Applying equation: s=ut+
2
1
at
2
(at h=40m)
⇒40=30t−
2
1
gt
2
⇒40=30t−5t
2
⇒t
2
−6t+8=0
As one root of the equation is given as: t
1
=2s
By product of roots t
1
t
2
=8
⇒t
2
=4s. (when the ball reaches the same height again)
Firstly we will calculate u from
T=2u/g
We got u:
2s=2u/9.8
u=9.8m/s
Now we know maximum height= u2/g
So substituting u=9.8m/s in it,
We get,
H= 9.8*9.8/9.8
H=9.8 m
Thank me later:)