A stone is thrown straight upward and it rises to a maximum...

A stone is thrown straight upward and it rises to a maximum height of $20 \mathrm{~m}$. With what speed was it thrown? Take up as the positive $y$ -direction. The stone's velocity is zero at the top of its path. Then $v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}$. (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use $v_{s}^{2}=v_{i}^{2}+2 a y$ to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$ Alternative Method You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, $y_{p}=-v_{i}^{2} / 2 s$, and so $v_{i}^{2}=-2 g_{y}$, or $v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20$ and $u_{i}=19.8 \mathrm{~m} / \mathrm{s}$, or to two significant figures, $u_{i}=20 \mathrm{~m} / \mathrm{s}$.

A stone is thrown straight upward and it rises to a maximum height of $20 \mathrm{~m}$. With what speed was it thrown?

Take up as the positive $y$ -direction. The stone's velocity is zero at the top of its path. Then $v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}$. (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use $v_{s}^{2}=v_{i}^{2}+2 a y$ to find
$$
v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s}
$$
Alternative Method
You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, $y_{p}=-v_{i}^{2} / 2 s$, and so $v_{i}^{2}=-2 g_{y}$, or $v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20$ and $u_{i}=19.8 \mathrm{~m} / \mathrm{s}$, or to two significant figures, $u_{i}=20 \mathrm{~m} / \mathrm{s}$.

Key Concepts

Kinematic Equations for Uniformly Accelerated Motion

These equations relate the variables of displacement, velocity, acceleration, and time for objects moving with constant acceleration. They are foundational in solving problems involving projectile motion, free-fall, and any motion under a constant force, allowing one to calculate unknown variables if some initial conditions are known.

Projectile Motion and Maximum Height

In projectile motion, the maximum height is reached when the vertical component of velocity becomes zero. This concept is critical because it allows one to set up equations where the final velocity at the peak is zero, thereby connecting the initial velocity, acceleration due to gravity, and the displacement (height reached).

Acceleration Due to Gravity

Gravity provides a constant acceleration that acts downward on objects near the Earth's surface. This acceleration is approximately 9.81 m/s². In kinematic calculations, its direction must be carefully considered by assigning a proper sign based on the chosen coordinate system.

Sign Conventions in Kinematics

Assigning appropriate positive and negative signs to variables such as displacement, velocity, and acceleration is crucial. Consistent sign conventions ensure that the physical meanings are correctly represented, such as treating upward motion as positive and recognizing that the gravitational acceleration is negative when upward is chosen as the positive direction.

Alternative Methods for Verification

Sometimes, different kinematic or energy conservation approaches can be used to verify the results. By applying various methods that rely on the same underlying physics concepts, one can cross-check the consistency and accuracy of calculated quantities such as the initial speed or maximum height of a projectile.

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