Question

ground.
7. A ball is thrown upwards with a velocity of 80m/s at an angle of 30° to the horizontal.
Find its velocity after one second.​

Answer 1

Solution :-

Given ,

• Velocity of the ball = 80 m/s
• Angle = 30°

We need to find ,

• Velocity after one second = ?

Firstly finding horizontal component velocity

• V = ucosθ

Substituting known values we have,

➵ V = 80 × cos30°

➵ V = 80 × √3/2

➵ V = 40 × √3

➵ V = 40√3

Now finding vertical component of Velocity ,

• V = usinθ - gt

➸ V = 80 × sin30° - 10 × 1

➸ V = 80 × 1/2 - 10

➸ V = 40 - 10

➸ V = 30

Now , Velocity vector :-

⇒ V = 40√3 i + 30j

⇒ V = √4800 + 900

⇒ V = 10√57 m/s

Hence , velocity = 10√57 m/s

Answer 2

Answer:

Given :-

• A ball is thrown upwards with a velocity of 80 m/s at an angle of 30° to the horizontal.

To Find :-

• What is the velocity after one second.

Formula Used :-

❶ Horizontal components of velocity,

$\boxed{\bold{\large{V\: =\: ucos{\theta}}}}$

❷ Vertical component of velocity,

$\boxed{\bold{\large{V\: =\: usin{\theta}\: -\: gt}}}$

❸ Velocity vector formula,

$\boxed{\bold{\large{V\: =\: ucos{\theta} +\: usin{\theta}\: -\: gt}}}$

Solution :-

Given :

• Velocity of the ball = 80 m/s
• Angle at an horizontal = 30°

❶ First, we have to find the horizontal component of velocity,

⇒ V = 80 × cos 30°

⇒ V = 80 × $\dfrac{\sqrt{3}}{2}$ [cos 30° = $\dfrac{\sqrt{3}}{2}$ ]

➠ V = $40\sqrt{3}$

❷ Again, we have to find the velocity component of velocity,

⇒ V = 80 × $\dfrac{1}{2}$ - 10(1) [sin 30° = $\dfrac{1}{2}$ ]

⇒ V = 40 - 10

➠ V = 30

❸ Last, we have to find the velocity vector formula,

⇒ V = 40$\sqrt{3}$ i + 30 j

⇒ V = $\sqrt{4800 + 900}$

➥ V = $10\sqrt{57}$

$\therefore$ The velocity after one second is $10\sqrt{57}$ .