Question

The equation of a path of a projectile which
is a projected with a velocity (V) at an angle
o with horizontal is cx and y are horizontal
and vertical displacement respectively)
2
y = x(tan)
gx​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

D-Q

Explanation:

Comparing given equation, y = ax - bx 2 with

the equation of projectile motion y = x tan θ - 2u 2 cos 2 θgx 2 we get, tanθ = a---(i) and

2 cos θ = b----(ii) ∴ 2u 2 gsec 2 θ

= b or 2u 2

g(1+tan

2

θ)

=b

or u

2

=

b

g(1+tan

2

θ)

=

2b

g(1+a

2

)

(Using (i))

A−s

(B) Horizontal range, R =

g

u

2

sin

2

θ

=

g

u

2

sinθcosθ

R=

g

u

2

cos

2

θ

X tanθ= a/b (Using (i) and (ii))

B−p

(C) Maximum height, H =

g

u

2

sin

2

θ

H=

2g

u

2

cos

2

θ

X tan

2

θ=

4g

u

2

cos

2

θ

X tan

2

θ=

4b

a

2

C−r (Using (i) and (ii))

(D) From eqn. (ii),

u

2

cos

2

θ= g/2b or ucosθ=

2b

g

-----(iii)

Time of flight=

g

2usinθ

=

g

2ucosθ

X tanθ

=

g

2

2b

g

Xa= a

bg

2

(Using (i) and (iii))

D−q