Question

a ball is thrown upward with speed 10m/s^1 from the top of the tower reaches the ground with a speed 20m/s^1 . the height if the tower is​

Answer 1

V^2 = U^2 + 2AS

v = 20 m/s

u = 0 m/s (velocity at highest point)

It implies,

400 = 0+ (2 * 10 * s)

400 = 20s

s = 20m (total distance covered by rock while going down from its highest point)

V^2 = U^2 + 2AS

0 = (10^2) + (2*-10*s)

100 = 20s

s = 5 (distance covered by stone when thrown up with an intital velocity of 10m/s)

Therefore tower is 20-5 = 15m tall.

Answer 2

A ball is thrown upward with speed 10m/s from the top of the tower reaches the ground with a speed 20m/s.

We have to find the height of the tower.

From above data we have; initial velocity is 10 m/s and final velocity is 20 m/s.

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

We have value of v and u. And we know that acceleration due to gravity is 9.8 approx. 10 m/s². So,

Using the Third Equation Of Motion:

v² - u² = 2as

Substitute the known values,

→ (20)² - (10)² = 2(10)s

→ 400 - 100 = 20s

→ 30 = 2s

Divide by 2 on both sides,

→ 30/2 = 2s/2

→ 15 = s

Therefore, the height of the tower is 15 m.