Question

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​

Answer 1

Answer:

√(M²+ 25g² - 10MgSinα)  - M

Explanation:

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 2

Answer:

the Answer is 50m/s

Explanation:

the velocity in horizontal direction remains same so the change is 0

the vertical velocity changes

$V_{5}$ = usin∅ - gt

$V_{o}$ = usin∅

change in velocity = $V_{o}$ - $V_{5}$ = gt = 10(5) = 50 m/s