Question

14.
A body projected at angle 'O' with the horizontal. If remains for a time T in air. Its horizontal range is R,
then tan =
(A) gT/2R
(B) g'T/2R
(C) gT?/2R
(D) GT/2R​

Answer 1

ANSWER

(a)

Consider that the initial velocity of projectile in X and Y direction be v

0x

and v

0y

respectively.

Now, as the particle reaches the point P during its motion, consider v

x

and v

y

be the velocity in X and Y directions respectively.

Using the forst equation of motion for the particle to reach the point P.

In vertical direction:-

v

y

=v

0y

=gt

And in horizontal direction:=

v

x

=v

0x

From the above equation, we can obtain the value of θ

tanθ=

v

x

v

y

=

v

0x

(v

0y

−gt)

θ=tan

−1

v

0x

(v

0y

−gt)

(b)

The maximum height of projectile is given by,

h

m

=

2g

u

0

2

sin

2

θ

...(i)

The horizontal range of the projectile is given by,

R=

g

u

0

2

sin2θ

... (ii)

Taking the ratio of the maximum height and range of projectile:

R

h

m

=

2sin

2

2θ

sin2θ

=

2×2sinθcosθ

Sinθ×sinθ

=

4Cosθ

sinθ

=

4

tanθ

tanθ=(4h

m

/R)

θ=tan

−1

(4h

m

/R)