Question

A ball is dropped from rest at the top of a building that is 320 m tall. Assuming no air resistance, what is its speed when it strikes the ground? Take acceleration due to gravity, g, as 9.81 m s-2. Give your answer to two significant figures and without a unit.

Answer 1

Answer:

g = 9.81 m/s²

h = 320 m

v = √(2gh)

v = √(2 × 9.81 × 320)

v = 79.24 m/s

Answer 2

Answer:

$\boxed{\mathfrak{Final \: velocity \: (v) \approx 79.24 \: m/s}}$

Given:

Initial velocity (u) = 0 m/s

Height of the building (h) = 320 m

Acceleration due to gravity (g) = 9.81 m/s²

To Find:

Final velocity (v)

Explanation:

As there is no air resistance acting on the ball so it's the case of free fall.

By using $\sf 3^{rd}$ equation of motion we can directly calculate the final velocity.

$\sf \star \: 3^{rd} \: equation \: of \: motion : \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2gh}}$

By substituting values of u, g & h in the equation we get:

$\sf \implies {v}^{2} = {(0)}^{2} + 2 \times 9.81 \times 320 \\ \\ \sf \implies {v}^{2} = 2 \times 9.81 \times 320 \\ \\ \sf \implies {v}^{2} = 6278.4 \\ \\ \sf \implies v = \sqrt{6278.4} \\ \\ \sf \implies v \approx 79.24 \: m/s$

$\therefore$

$\sf Final \: velocity \: (v) \approx 79.24 \: m/s$