Question

a body thrown vertically upward reaches its maximum height in time T the body is at a height h which is less than maximum height at two instance of time T1 and t2 hence product of T1 T2 is given by

Answer 1

Answer:

2h/g

Explanation:

Let the maximum height be H and u be initial velocity

H = uT - gT²/2

0 =  u - gT , T = u/g

H = gT²/2 = u²/2g

Now

h = ut - gt²/2

gt²/2 - ut + h = 0

t² - (2u/g)t + 2h/g = 0

t² - 2Tt + 2h/g = 0

The product of t1 & t2 will be the product of the two roots of the above quadratic equation

t1t2 = T1T2 = 2h/g

Another solution :

t1,2 = T ±√(T²-2h/g)

t1t2 = { T+ (√(T²-2h/g)}{T+ (√(T²-2h/g)}

t1t2 = T² - T² + 2h/g

t1t2 = 2h/g

Answer 2

Answer:

t1×t2 = 2h/g

Explanation:

pls see the attachment... pls understand one thing t2 = t1 + 2(t3)