Question

a ball is thrown with 5m/ s velocity at angle of projection 45⁰ find 1,time to flight 2, maximum height 3, horizontal range​

Answer 1

Given :

• Speed of projection = 5 m/s
• Angle of projection =45 ⁰

To find :

• Time of light
• Maximum height
• Horizontal range

How to solve :

• Simply substitute the given values  in the formula to get the answer .

Solution :

(1) Time of light

As per the formula ,

⇒ T = 2 u sin Ф / g

Now let's substitute the given values ,

⇒ T = 2 x 5 x sin 45 / 10

⇒ T = 2 x 5 x 1 / 10 √2

⇒ T = 1 / √2

⇒ T = 0.7 s

(2) Maximum Height

⇒ H = u² sin²Ф / 2g

Now let's substitute the given values ,

⇒ H = 5² x ( sin 45 )² / 2g

⇒ H = 25 x 1 / 2 x 2 x 10

⇒ H = 25 / 40

⇒ H = 0.625 m

(3) Horizontal range

⇒ R = u² sin 2Ф / g

we know that ,

• sin2Ф = sin Ф . cos Ф

Now let's substitute the given values,

⇒ R = u ² sin Ф . cos Ф / g

⇒ R = 5 x 5 x sin 45 . sin 45 / 10

⇒ R = 25 x 1 / 2 x 10

⇒ R = 25 / 20

⇒ R = 1.25 m