Question

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor.

Calculate the velocity just before the tennis ball strikes the floor.
1. v_{{before}} = sqrt{2gh_{{initial}}}
2. v_{{before}} = frac{2gh_{{initial}}}{t_{{fall}}}
3. v_{{before}} = frac{g ⋅ t_{{fall}}}{2h_{{initial}}}
4. v_{{before}} = frac{2h_{{initial}}}{gt_{{fall}}}

Answer 1

The velocity of a soft tennis ball just before it strikes the floor from a height of 1.50 m is calculated using the energy conservation formula, resulting in a velocity of approximately 5.42 m/s.

Step-by-step explanation:

To calculate the velocity of a soft tennis ball just before it strikes the floor from a height of 1.50 m, we can use the formula derived from the conservation of energy principle: v_{before} = \sqrt{2gh_{initial}}, where g is the acceleration due to gravity (9.81 m/s2), and h_{initial} is the initial height.

To find the velocity just before it hits the floor, plug in 1.50 m for h_{initial}:

v_{before} = \sqrt{2 \times 9.81 m/s2 \times 1.50 m}

Solving this, we find that the velocity before impact is:

v_{before} = \sqrt{29.43 m2/s2}

v_{before} = 5.42 m/s (rounded to two decimal places)

This is the correct option for part (a) of the question. To address parts (b), (c), and (d), additional calculations using kinematic formulas and dynamics would be necessary.