Question

A particle is projected from the ground with a velocity of 40 m/s at an angle of 30° with the horizontal The
ground there is a horizontal plane.
The speed of particle at the highest pointis
(A) zero
(B) 2013 m/s
(C) 20 m/s
(D) none​

Answer 1

➠Given :

• Initial velocity (u) = 40m/s
• Angle of projection (Φ) = 30°

➠To Find :

• Velocity of projectile at highest point.

➠SoluTion :

➨ A body is said to be projectilr if i is projected into space with some initial velocity and then it continuoes to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration equal to g.

➨ At maximum height,

Vertical velocity = zero

Horizontal velocity = ucosΦ

⇒ Vx = u cosΦ

⇒ Vx = 40 × cos30°

⇒ Vx = 40 × √3/2

⇒ Vx = 20√3 m/s

Answer 2

$\blue{\bold{\underline{\underline{Answer:-}}}}$Given :   u=40m/s                     ay=−g=−10m/s2

∴   ux=ucos60o=20m/s             and      uy=usin60o=34.64m/s

x direction :    ax=0              ⟹Vx=ux=20m/s

Also    VxVy=tan45o=1                 ⟹Vy=Vx=20m/s

y direction :            Vy=uy+ayt

Upward motion :        20=34.64−10t1                     ⟹t1=1.464  s

Downward motion :        −20=34.64−10t2             ⟹t2=5.464  s