Question

a body is thrown at some angle from the ground the magnitude of change in velocity of body in 5s if it is still in air g=10m/s*2
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Answer 1

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√(M²+ 25g² - 10MgSinα) - M

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A body is thrown at some angle from the ground the magnitude of change in velocity of a body in 5 second if it is still in air is

Let say body is thrown at angle of α with Magnitude M

Horizontal Velocity = MCosα

Vertical Velocity = MSinα

Horizontal Velocity remains same as acceleration = 0

in Vertical Velocity  , a =  - g  ( acceleration due to gravity)

Vertical Velocity after 5 sec =   M Sinα - 5g

Velocity Magnitude after 5 sec  =  √(MCosα)² + (M Sinα - 5g)²

= √M²Cos²α + M²Sin²α + 25g² - 10MgSinα

= √M²Cos²α + M²Sin²α + 25g² - 10MgSinα

= √(M²+ 25g² - 10MgSinα)

Change in magnitude

= √(M²+ 25g² - 10MgSinα)  - M

Answer:-

(1) 50 m/s

Explanation:

g=V-u/t

10=v-0/5

: u=0, because the object starts from rest position

v=10×5

v=50 m/s