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Chapter 2: Problem 59

A stone is thrown upward, from ground level, with an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\). a) What is the velocity of the stone after 0.50 s? b) How high above ground level is the stone after 0.50 s?

Short Answer

Expert verified
Answer: The velocity of the stone after 0.50 s is 5.095 m/s, and its height above ground level is 3.77375 m.

Step by step solution

01

Identify the relevant variables and kinematic formula.

We need to find the final velocity (v) of the stone after 0.50 s with an initial velocity (u) of 10.0 m/s. The acceleration due to gravity (a) is -9.81 m/s², and the time (t) is 0.50 s. The kinematic formula we need is: v = u + at
02

Plug in the known values and solve for the final velocity.

Input the given values into the kinematic formula: v = 10.0 m/s + (-9.81 m/s²)(0.50 s) Now, solve for v: v = 10.0 m/s - 4.905 m/s v = 5.095 m/s So, after 0.50 s, the velocity of the stone is 5.095 m/s. b) Finding the height above ground level after 0.50 s.
03

Identify the relevant variables and kinematic formula.

We need to find the displacement (s) of the stone from the ground after 0.50 s with an initial velocity (u) of 10.0 m/s. The acceleration due to gravity (a) is -9.81 m/s², and the time (t) is 0.50 s. The kinematic formula we need is: s = ut + 0.5at²
04

Plug in the known values and solve for the displacement.

Input the given values into the kinematic formula: s = (10.0 m/s)(0.50 s) + 0.5(-9.81 m/s²)(0.50 s)² Now, solve for s: s = 5.0 m - 1.22625 m s = 3.77375 m So, after 0.50 s, the stone is 3.77375 m above the ground level.

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