Question 2 Not yet answered A ball with a diameter of your last digit of your...

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Answer 1

Answer #1

Third equation of motion,

v^2 = u^2 + 2as

Where, v : final velocity

u: initial velocity

a: acceleration

s : distance travelled

Apply this equation of motion in vertical direction.

Initial velocity, u = 6m/s

Acceleration due to gravity, a = -g = -10 m/s² (-ve sign because it is downward)

Velocity will be zero at maximum height so v = 0, s = H (maximum height)

$\\0 = 6^2 + 2(-10)H\\ \\ 20H = 36\\\\ H = \frac{36}{20} = \textbf{1.8 m}$

Correct answer is (h) 1.8 m

Note: Maximum height does not depend on mass or diameter of the ball.

Answer 2

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