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Chapter 2: Problem 47

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. Grant Hill jumps \(1.3 \mathrm{m}\) straight up into the air to slamdunk a basketball into the net. With what speed did he leave the floor?

Short Answer

Expert verified
Answer: Grant Hill's initial speed was approximately 5.05 m/s.

Step by step solution

01

Determine the final velocity at the highest point

At the highest point of the jump, Grant Hill's final velocity will be 0 (v = 0)
02

Write down the given values and the kinematic equation

We have: - Final velocity, \(v = 0 \ \mathrm{m} / \mathrm{s}\) - \(g = 9.80 \ \mathrm{m} / \mathrm{s^2}\) - Displacement, \(s = 1.3 \ \mathrm{m}\) We'll use the kinematic equation: \(v^2 = u^2 + 2as\)
03

Substitute the values in the equation

Substitute the values in the equation: \(0^2 = u^2 + 2(- 9.80) (1.3)\)
04

Solve the equation for initial speed (u)

Solving for \(u\): \(u^2 = 0 - 2(-9.80)(1.3)\) \(u^2 = 25.48\) \(u = \sqrt{25.48}\) (since we're looking for the positive value of speed) \(u \approx 5.05 \ \mathrm{m} / \mathrm{s}\) So, Grant Hill left the floor with an initial speed of approximately \(5.05 \ \mathrm{m} / \mathrm{s}\).

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