Question

a stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u.the height of the tower is

Answer 1

v² - u² = 2aS

H = (v² - u²) / (2g)

H = (9u² - u²) / (2g)

H = 4u²/g

Height of the tower is 4u²/g

Answer 2

Answer:

The “height of the tower” is $\bold{\frac{4 u^{2}}{g}}$.

Solution:

The final and initial velocity of the stone has been provided as 3u and u respectively.

As the stone is falling down from upward, so the acceleration will be equal to “acceleration due to gravity”.

According to Newton’s second equation of motion:

$v^{2}-u^{2}=2 a s$

Here ‘v is the final velocity’, ‘u is the initial velocity’, ‘a is the acceleration’, and ‘s is the distance’.

So in the present case, v = 3u, u = u, a = g and s = h = height of the tower

$(3 u)^{2}-u^{2}=2 g h$

$9 u^{2}-u^{2}=2 g h$

$h=\frac{9 u^{2}-u^{2}}{2 g}$

$h=\frac{8 u^{2}}{2 g}$

$h=\frac{4 u^{2}}{g}$

Thus, height of the tower is $\bold{\frac{4 u^{2}}{g}}$