Question

a ball is projected with velocity of 10m/s at angle of 30with horizontal surface .the speed of the ball after 1 sec will be (use g=10​

Answer 1

Given :

➳ Initial velocity = 10m/s

➳ Angle of projection = 30°

➳ Acc. due to gravity = 10m/s²

To Find :

⟶ Speed of ball after 1s.

Concept :

➠ In projectile motion, Horizontal component of velocity remains constant throughout the motion. (No acceleration in horizontal direction)

➠ Vertical component of velocity changes continuously as gravitational acceleration acts in downward direction.

➠ Horizontal component of velocity after time t :

• Vx = u cosΦ

➠ Vertical component of velocity after time t :

• Vy = u sinΦ - gt

◈ Final velocity after time t is given by,

$\bigstar\:\boxed{\bf{V=\sqrt{{V_x}^2+{V_y}^2}}}$

Calculation :

$\leadsto\sf\:V=\sqrt{(u\cos\theta)^2+(u\sin\theta-gt)^2}$

$\leadsto\sf\:V=\sqrt{u^2+g^2t^2+(2u\sin\theta)gt}$

$\leadsto\sf\:V=\sqrt{(10)^2+(10)^2(1)^2+(2\times 10\times \sin30\degree)(10\times 1)}$

$\leadsto\sf\:V=\sqrt{100+100+(10)(10)}$

$\leadsto\sf\:V=\sqrt{300}$

$\leadsto\boxed{\bf{V=10\sqrt{3}}\:ms^{-1}}$