Question

A ball is projected at an angle of 37° from a
point. It manages to cross a wall which is at a
distance of 20 m from the point of throw. If the
range of the ball on the ground is 80 m, the
maximum possible height of the wall is
(1) 45 m
(2) 22.5 m
(3) 11.25 m
(4) 20 m

Answer 1

Given : x=40 m y=15 m

Using equation of trajectory y=xtanθ−

2u

2

cos

2

θ

gx

2

where θ=45

o

⟹ tan45=1 cos45=

2

1

Or 15=40×1−

2×u

2

×

2

1

10 ×40

2

Or

u

2

10×40

2

=25

⟹ u

2

=640

Maximum height attained H

max

=

2g

u

2

sin

2

θ

⟹ H

max

=

2×10

640×

2

1

=16 m