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
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)]