A football is kicked straight up into the air; it hits the ground 3.6 s later. A) What was the greatest height reached by the ball? Assume it is kicked from ground level. B) With what speed did it leave the kicker's foot?

To solve the question given above, we begin by calculating the time taken for the ball to get to the greatest height. This can be obtained as follow:

Time of flight (T) = 3.6 s

Time taken to reach the greatest height (t) =?

T = 2t

3.6 = 2 × t

Divide both side by 2

t = 3.6 / 2

t = 1.8 s

A. Determination of the greatest height.

Time taken to reach the greatest height (t) = 1.8 s

Acceleration due to gravity (g) = 9.8 m/s²

Greatest height (h) =?

h = ½gt²

h = ½ × 9.8 × 1.8²

h = 15.876 m

Therefore, the greatest height reached by the ball is 15.876 m

B. Determination of the initial velocity.

Greatest height (h) = 15.876 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at greatest height)

Initial velocity (u) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 15.876)

0 = u² – 311.1696

Collect like terms

0 + 311.1696 = u²

311.1696 = u²

Take the square root of both side

u = √311.1696

u = 17.64 m/s

Therefore, the ball left the kicker's foot with a speed of 17.64 m/s

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