Physics, asked by nandinikashiv23, 2 months ago

An object at height h is projected vertically upwards with speed u . Find its speed just before collision with the ground and also find its time of flight? ​

Answers

Answered by abhi569
3

Answer:

Speed = √(u² + 2gh)

Time = (u/g) [1 + √(1 + 2hg/u²)]  

Explanation:

At the highest point, object stops for a moment, to fall down (velocity = 0). Using eqⁿ of motion:

=> v = u + at     => 0 = u + (-g)t    => u/g = t,

where t is the time to reach highest point from the height h. If x is the length between tower and highest point,

v²= u² + 2aS  =>0 = u² + 2(-g)x  =>u²/2g = x

From the highest point to ground:

Initial velocity = 0  ; total height = x + h ;

=> v² = u² + 2aS  

=> v² = 0 + 2g(x + h)

=> v² = 2g(u²/2g + h)           [x = u²/2g]

=> v² = u² + 2gh

=> v = √(u² + 2gh)  

Using S = ut + ½ at²,let total time(down)=T

=> (x + h) = 0t + ½ g(T)²

=> (u²/2g + h) = ½ g(T)²

=> √(u² + 2gh)/g = T

Total time in air = t + T

= u/g + √(u² + 2gh)/g

= (u + √(u² + 2gh))/g

Hence, required speed = √(u² + 2gh) and required time = (u + √(u² + 2gh))/g

required time = (u/g) [1 + √(1 + 2hg/u²)]

__Using vectors___: a short method:

It is thrown with u upwards, it implies its velocity is -u downwards. Let the time to go down be T and final velocity is v.

=> v² = (-u)² + 2gh

=> v = √u² + 2gh  = velocity near ground.

Time to cover height h:

h = -uT + ½ gT²   => 0 = gT² - 2uT - 2h

Using quadratic formula,

T = (u ± √u² + 2gh)/g

T = (u/g) [1 + √(1 + 2gh/u²)]

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