Question

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

Question: A stone is thrown upwards with an initial velocity of 10 m/s. Calculate its velocity and the height above ground after 0.50 s. Answer: a) The velocity of the stone after 0.50 s is approximately 5.095 m/s. b) The height above ground level after 0.50 s is approximately 1.274 m.

Step-by-step solution

step 1: Find the velocity after 0.50 s

To find the velocity, plug in the given values into the equation \(v = v_0 - gt\). We have \(v_0 = 10\mathrm{~m/s}\) and \(t = 0.50\mathrm{~s}\). The gravitational constant \(g\) is approximately 9.81\(\mathrm{~m/s^2}\). We get: \(v = 10\mathrm{~m/s} - (9.81\mathrm{~m/s^2})(0.50\mathrm{~s})\) Calculating the expression, we find the velocity of the stone after 0.50 s: \(v \approx 5.095\mathrm{~m/s}\) (rounded to 3 decimal places)

step 2: Find the height above ground level after 0.50 s

To find the height, plug in the given values into the equation \(y = v_0t-\dfrac{1}{2}gt^2\). We have \(v_0 = 10\mathrm{~m/s}\) and \(t = 0.50\mathrm{~s}\). The gravitational constant \(g\) is approximately 9.81\(\mathrm{~m/s^2}\). We get: \(y = (10\mathrm{~m/s})(0.50\mathrm{~s})-\dfrac{1}{2}(9.81\mathrm{~m/s^2})(0.50\mathrm{~s})^2\) Calculating the expression, we find the height above ground level after 0.50 s: \(y \approx 1.274\mathrm{~m}\) (rounded to 3 decimal places) So, a) The velocity of the stone after 0.50 s is approximately 5.095 m/s. b) The height above ground level after 0.50 s is approximately 1.274 m.

Key concepts

Projectile Motion

Projectile motion is a form of motion experienced by an object that is thrown or projected into the air and is acted upon only by gravity and air resistance. In an idealised scenario, where air resistance is neglected, the only force acting on the projectile is gravity, which accelerates the object towards the ground at a constant rate. This motion is two-dimensional and can be analysed by separating it into two components: horizontal and vertical.

Let's take a stone thrown upward with an initial velocity – as in the exercise. The stone's upward path and downward fall constitute its projectile motion. The initial velocity, combined with the force of gravity, will determine how high the stone will rise, how long it will stay in the air, and where it will land. The path of a projectile is called its trajectory, which, in the absence of air resistance, forms a parabolic shape.

Gravitational Acceleration

Gravitational acceleration, often denoted as \(g\), is the acceleration of an object due to the force of gravity. On the surface of the Earth, this value is approximately \(9.81\ \mathrm{m/s^2}\) towards the center of the planet. This acceleration is what gives weight to objects and causes projectiles to eventually fall to the ground.

When we calculate the motion of objects like the stone in our exercise, we assume that gravitational acceleration is constant and acts downward. In our equation, \(v = v_0 - gt\), the term \(gt\) represents the change in velocity due to gravity. Over the time of half a second, gravity reduced the stone’s upward velocity by \(4.905\ \mathrm{m/s}\) (half of \(9.81\ \mathrm{m/s^2}\) times \(0.50\ \mathrm{s}\)), resulting in the final velocity.

Equations of Motion

The equations of motion describe the relationship between an object's displacement, initial velocity, constant acceleration, and the time taken. For an object under the influence of gravity, we use two key equations:

\(v = v_0 - gt\) gives us the final velocity (\(\v\)) of an object after a time (\(\t\)) has passed, with \(\v_0\) as the initial velocity and \(\g\) representing the acceleration due to gravity. In the example, after 0.50 seconds, we adjusted the initial velocity by subtracting the change in velocity caused by gravitational acceleration.

Equation \(\y = v_0t - \frac{1}{2}gt^2\) calculates the vertical displacement or height (\(\y\)) of the object after time (\(\t\)). It considers the initial vertical velocity (\(\v_0\)) and the downward pull of gravity over time. For the stone, we concluded that after 0.50 seconds, it reached a height of about 1.274 meters above ground level. Such equations are fundamental in predicting an object’s motion, as we saw in determining the stone's path in the exercise.