Question

A ball is dropped from the top a tower 40m high. What is its velocity when its is covered a distance of 20m?what would be its hits the ground? find the velocity just before it hits the ground
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Answer 1

Answer:

V = 28.01 m/s

Explanation:

The total height of the tower is 40 m.

Lets total time be = t

Initial velocity u= 0

Case 1: whats the velocity when the body at 20 m. High

From the formula

V^2 - u^2 =2 * a* s

V^2 - 0^2 = 2*9.81* 20

V^2 = 392.4

V= √392.4

V = 19.809 m/s

Case2 :velocity when the ball hits the ground

From

V^2 - u^2 =2 * a* s

V^2 - 0^2 = 2 *9.81* 40

V^2 = 784.8

V=√784.8

V = 28.01 m/s

Answer 2

$\bf{\underline{\underline{Question:-}}}$

A ball is dropped from the top of tower 40m high. What is its velocity when it has covered 20m? What would its velocity when it hits the ground?

$\bf{\underline{\underline{Solution:-}}}$

• The total height of the tower is 40 m.

• Lets total time be = t

• Initial velocity u= 0

Case 1:

• whats the velocity when the body at 20 m. High

From the formula :-

→ V² - u² =2 × a × s

→ V² - 0² = 2 × 9.81 × 20

→ V² = 392.4

→ V= $\sf \sqrt{392.4}$

→ V = 19.809 m/s

Case2 :

• velocity when the ball hits the ground

From the formula :-

→ V² - u² =2 × a × s

→ V² - 0² = 2 × 9.81 × 40

→ V² = 784.8

→ V = $\sf \sqrt{784.8}$

→ V = 28.01 m/s

$\bf{\underline{\underline{Hence:-}}}$

• The velocity of the body at 20m high is 19.809 m/sec
• velocity when the ball hits the ground is 28.01 m/sec