A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m? b. How long will it be in the air?

A. 29.7 m/s

B. 6.06 s

Explanation:

From the question given above, the following data were obtained:

Maximum height (h) = 45 m

A. Determination of the initial velocity (u)

Maximum height (h) = 45 m

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

Final velocity (v) = 0 m/s (at maximum height)

Initial velocity (u) =.?

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

0² = u² – (2 × 9.8 × 45)

0 = u² – 882

Collect like terms

0 + 882 = u²

882 = u²

Take the square root of both side

u = √882

u = 29.7 m/s

Therefore, the ball must be thrown with a speed of 29.7 m/s.

B. Determination of the time spent by the ball in the air.

We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:

Maximum height (h) = 45 m

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

Time (t) to reach the maximum height =?

h = ½gt²

45 = ½ × 9. 8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:

Time (t) to reach the maximum height = 3.03 s

Time (T) spent by the ball in the air =?

T = 2t

T = 2 × 3.03

T = 6.06 s

Therefore, the ball spent 6.06 s in the air.

momentum of soft ball three times less than baskey ball

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