Question

10. A body is projected with an initial velocity of
10/3m s making an angle of 30° with the
horizontal. The velocity of the particle at the highest
point of the trajectory is :
(a) 15 m s-1
(b) 53 m s -1
(c) o
(d) 103 m s -1
Torino

Answer 1

Question :

A body is projected with an initial velocity of  10√3 m/s making an angle of 30° with the  horizontal. The velocity of the particle at the highest  point of the trajectory is :

(a) 15 m/s

(b) 53 m/s

(c) 0 m/s

(d) 10√3 m/s

Solution :

Projected velocity of the body , u = 10√3 m/s

So , Initial velocity , uᵧ = u.sinθ= 10√3 × ¹/₂

⇒ uᵧ = 5√3 m/s

Angle made with horizontal , θ = 30°

Time taken to reach highest point , t = $\sf \dfrac{usin \theta}{g}$

⇒ t = $\sf \dfrac{10 \sqrt{3}. sin30}{g}$

Final velocity of the particle at highest point , vᵧ = ? m/s

Apply 1st equation of motion ,

$:\implies \sf v_y=u_y-gt$

g is -ve , as it is moving upwards i.e., against the gravity

At highest point vᵧ = v ,

$:\implies \sf v=5\sqrt{3}-g\bigg(\dfrac{10\sqrt{3}.sin30}{g} \bigg)$

$:\implies \sf v=5 \sqrt{3}-10 \sqrt{3}. \dfrac{1}{2}$

$:\implies \sf v=5 \sqrt{3} -5 \sqrt{3}$

$:\implies \bf v=0\ m/s\ \; \green{\bigstar}$

So , Magnitude of the velocity of the particle at highest point will be 0 m/s

• Option (c)