Question

9) If y= ax=b2c? is the eqn for projectile motion
Find initial velocity, angle of projection, Time
of flight, max height,horizontal range​

Answer 1

Comparing given equation, y = ax - bx

2

with

the equation of projectile motion y = x tan θ -

2u

2

cos

2

θ

gx

2

,

we get, tanθ = a---(i) and

2u

2

cos

2

θ

g

= 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−