Question

a stone thrown vertically upwards from the top of a tower with an initial velocity u reaches the ground with a velocity 3u the height of the tower is

Answer 1

H = (v² - u²) / (2g)

H = (9u² - u²) / (2g)

H = 8u² / (2g)

H = 4u²/g

Height of the tower is 4u²/g

Answer 2

Answer:

Height of the tower is in terms of initial velocity as $\mathrm{H}=\frac{4 \mathrm{u}^{2}}{\mathrm{g}}$

Given:

Initial velocity = u

Acceleration, a = g

Final velocity, v = 3u

Solution:

The stone thrown moves with initial velocity and acceleration as given, the height of the tower is calculated using equation of motion, we have,

$\mathrm{H}=\mathrm{v}^{2}-\frac{\mathrm{u}^{2}}{2 \mathrm{g}}$

$\mathrm{H}=\mathrm{v}^{2}-\frac{\mathrm{u}^{2}}{2 \mathrm{g}}$

$\mathrm{H}=(3 \mathrm{u})^{2}-\frac{(\mathrm{u})^{2}}{2 \mathrm{g}}$

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

$\mathrm{H}=\frac{8 \mathrm{u}^{2}}{2 \mathrm{g}}$

$\mathrm{H}=\frac{4 \mathrm{u}^{2}}{\mathrm{g}}$