Question

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. speed = acceleration = choose direction... choose direction...

Explanation:

Step1: Analyze the motion of the rock

When the rock is thrown straight up, at the moment it leaves the person's hand, it has an initial velocity. As it moves upward, the only force acting on it (ignoring air - drag) is the force of gravity.

Step2: Determine the acceleration

According to Newton's second law \(F = ma\), and the force due to gravity \(F=-mg\) (taking the upward direction as positive). So, \(a=\frac{F}{m}=\frac{-mg}{m}=-g\). The value of \(g = 9.8\ m/s^{2}\). The acceleration is constant throughout the motion (ignoring air - drag) and is equal to the acceleration due to gravity.

Step3: Determine the speed at the peak

At the peak of its motion, the velocity of the rock is \(0\ m/s\). But the question asks for the speed at the moment it leaves the hand. We need to convert the given speed from miles per hour to meters per second.
We know that \(1\ mile = 1609.34\ m\) and \(1\ hour=3600\ s\).
If \(v = 84\ mph\), then \(v=84\times\frac{1609.34}{3600}\ m/s\approx37.57\ m/s\)

Answer:

Speed \(= 37.57\ m/s\), Acceleration \(=- 9.8\ m/s^{2}\) (direction: downward)