Question

Prove that if a body is thrown vertically upwards, then the time of ascent is equal to the time of descent.

Answer 1

Let,

u = Initial velocity

v = Final velocity

h = Height attained

a = acceleration

t = time

Ascent:

Ball thrown upward with velocity u. So, v = 0 at maximum height h. Also, a = -g

Now, using v = u + at

0 = u - gt_a

t_a = \(\frac{u}{g}\).......................1

Where t_a = time of ascent

Also, by using V² = U² + 2as

0 = u² - 2gh

u = \(\sqrt {2gh}\).............2

From equations 1 and 2, we get,

t_a = \(\sqrt \frac{2h}{g}\)..........A

Descent:

Ball comes down from height h. So, u = 0 and a = g.

So, using v = u + at

v = gt_d ….......3

Where t_d = Time of descent

Now, using v² = u² + 2as

v² = 0 + 2gh

v = \(\sqrt {2gh}\)................4

From equations 3 and 4, we see that,

t_d = \(\sqrt \frac{2h}{g}\)..........B

From equations A and B, we see that,

t_a = t_d