Question

A cricketer hits a ball with a velocity 25 m/s
at 60° above the horizontal.How far above
the ground it passes over a fieder 50 m from
the bat (assume the ball is struck very close
to the ground)
ans : 5.2 m
describe the steps...​

Answer 1

Answer:

Equation of trajectory when a projectile is thrown with velocity u at an angle θ with the horizontal is

y=xtanθ−

2

1

u

2

cos

2

θ

gx

2

y=50tan60

0

−

2

1

×

25×25×[

2

1

]

2

9.8×50×50

=50

3

=9.8×8

=86.60−78.4=8.2m

Answer By. shübhãm

follow me inst abhi.shubham.46

Answer 2

Answer:

hello

Explanation:

at 60° above the horizontal.How far above

the ground it passes over a fieder 50 m from

the bat (assume the ball is struck very close

to the ground)

ans : 5.2 m

describe the steps...

1