Question

free fall
the state an object is in when gravity is the only force acting on it. near the
surface of a planetary body an object in free fall will have a constant
acceleration. this acceleration is referred to as the \free fall acceleration\. on
earth, the free fall acceleration is 9.8 m/s² down.
problem
a person throws a rock straight up into the air. at the moment it leaves the persons hand it is going 84 mph. when the rock reaches its peak, how fast is it going and what is the magnitude and direction of
its acceleration? ignore air drag. express your answer using appropriate mks units.
speed =
acceleration =
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Explanation:

Step1: Convert speed units

We know that \(1\ mph=\frac{0.44704\ m}{1\ s}\).
Given initial speed \(v_0 = 84\ mph\), so \(v_0=84\times0.44704\ m/s\approx37.55\ m/s\).
At the peak of the motion, the vertical component of velocity \(v = 0\ m/s\).

Step2: Analyze acceleration

When an object is in free - fall (ignoring air drag), the only force acting on it is gravity.
The acceleration due to gravity near the Earth's surface \(a=-g=- 9.8\ m/s^{2}\) (negative because it acts in the downward direction).

Step3: Find the speed at the peak

At the peak of the motion (the highest point of the rock's trajectory), the vertical velocity of the rock is \(v = 0\ m/s\).

Answer:

Speed \(=0\ m/s\), Acceleration \(=-9.8\ m/s^{2}\) (downward)