Question

Answer this question ​

Answer 1

ANSWER :

OPTION A

GIVEN :

• Velocity of projection =20√3

• angle with the horizontal =60°

SOLUTION :

(a)Time of flight

As per the formula ,

T =2 u sin x / g

T =2×20√3 sin 60° /10

T = 40√3×√3/20

T=2×3

T=6 sec

(b) Range

Range is the horizontal distance covered by the projectile during it's motion

R = u² sin 2 x / g

R = u² 2 sin x . cos x /g

R = 40×3 sin 60 .cos 60/g

R = 40×3×√3×1/2×2×10

R=120√3/40

R =3√3 m

(3) Max Height

H = u ² sin ² x / 2 g

H = 40×3( √3)²/(2)² ×2 g

H = 120×3 /8g

H =360/8g

H =4.5 m

(d) At max height the vertical component of velocity becomes zero

Hence ,

U x = u cos x

U x = 20√3 cos 60°

U x = 20√3 /2

U x = 10 √ 3 m/s²