Q.5.A cricketer throws a ball at an angle of 30° with the horizontal having velocity 30ms⁻¹. The maximum height attained by the ball is(Take g = 10 ms⁻²) *
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Answer:
Here θ=30
o
,u=30ms
−1
a. The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and descending is same for a projectile without air resistance. the time to reach the highest point T
t
H
=
2
T
=
g
usinθ
=
10
30
×sin30
o
=1.5s
b. The maximum height reached is
2g
u
2
sin
2
θ
=
2g
(30
2
)×(sin30
o
)
2
=
2×10×4
900
=11.25m
c. Horizontal range =
g
u
2
sin2θ
=
10
(30)
2
sin2(30
o
)
=
20
900
3
=45
3
m
d. The time for which thc ball is in air is same as its time of eight, i.e.,
g
2usinθ
=
10
2×30×sin30
o
=3s
Explanation:
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