05Classical Mechanics

Vertical throws

A body thrown straight up slows down at the rate \(g\), stops for an instant at its highest point and comes back down speeding up at the same rate; one thrown down simply speeds up from the start. Both are variants of the invariant with \(a = -g\) along an axis pointing up. Choose the direction of the throw, the launch height if it leaves from a balcony or a roof, and the condition of the statement (maximum height, return to the hand, arrival at the ground); the laboratory solves it, draws the flight and tells apart the two instants at which the body passes through the same height.

Laboratory

\(y(t)\) is a parabola opening downwards, \(v(t)\) a straight line of slope \(-g\) that crosses zero at the top, and \(a(t)\) the constant \(-g\).

The invariant

With the \(y\) axis pointing up, the acceleration is \(-g\) and every velocity carries its sign: positive while the body rises, negative while it falls. The five equations of the invariant become the ones below, each without one of the quantities.

Step-by-step solution

Notable events

The instants every vertical throw has: the top, the return to the launch point and the arrival at the ground.

Two bodies on the same vertical

Two bodies thrown or dropped along the same vertical meet when \(y_A(t) = y_B(t)\). Both feel the same acceleration \(-g\), so the \(t^2\) terms cancel and the meeting condition becomes linear: relative to each other they move at constant velocity.

Theory

The vertical throw as a variant

In the course notes the throw upwards comes from the invariant by applying its conditions: the acceleration is \(g\) and it opposes the motion, so

\[V(t) = V_o - g\,t,\qquad V(t)^2 = V_o^2 - 2g\,H,\qquad H(t) = V_o\,t - \frac{g\,t^2}{2},\]

while for a throw downwards the velocity and \(g\) point the same way and \(V = V_o + gt\), \(V^2 = V_o^2 + 2gH\), \(H = V_o t + \tfrac{gt^2}{2}\). Here \(g = \Delta V/\Delta t = 9.81\ \tfrac{\text{m/s}}{\text{s}}\): the velocity changes 9.81 m/s every second.

One axis and one sign convention

The safest way to handle the whole flight is Resnick’s: a single \(y\) axis pointing up and \(a = -g\) with \(g > 0\). Then \(v = v_0 - gt\) and \(y = y_0 + v_0 t - \tfrac12 g t^2\) describe the rise, the top and the fall without changing anything: the velocity is positive while it rises, zero at the top and negative while it falls.

Rise and fall are symmetric

The top is reached when \(v = 0\), at \(t_s = v_0/g\), and the maximum height above the launch point is \(H = \dfrac{v_0^2}{2g}\). Setting \(\Delta y = 0\) gives \(t = 2v_0/g\): the body takes as long to come down as to go up, and it comes back to the hand with the same speed, \(v = -v_0\). At every height it passes twice, with velocities equal in magnitude and opposite in sign.

At the top, v = 0 but a ≠ 0

A common mistake is to think that the acceleration vanishes at the highest point. The velocity is zero there for an instant, but it is still changing at the rate \(-g\): one instant before it was positive and one instant later it is negative. If the acceleration were zero at the top the body would stay there.

Throws from a height: two roots

When the body leaves from a height \(y_0\) and we ask when it reaches the ground, \(0 = y_0 + v_0 t - \tfrac12 g t^2\) is a quadratic with one positive and one negative root. As the course notes say, we keep the positive root, since there are no negative times; the negative one corresponds to the instant at which a body following the same parabola would have left the ground.

Two answers for one height

Asking when the body is 8 m above the launch point gives two times, one on the way up and one on the way down, and \(v^2 = v_0^2 - 2g\Delta y\) gives two velocities of opposite sign. Both are physically valid, and the laboratory shows them as two solutions. Only if the height exceeds \(v_0^2/2g\) is there no solution: the body never gets there.

Problem bank

Problems adapted from Serway (chapters 2, 4, 5 and 6), Resnick and the course notes. Load any of them into the laboratory: the data fill the panel and the solver does the rest, so you can then change any value and see how the answer reacts.

  1. A baseball hit straight up

    Serway 2.41 · class notes

    A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes \(3.00\ \text{s}\) for the ball to reach its maximum height. Find (a) the initial velocity of the ball and (b) the height it reaches.

  2. From a balcony

    Course workshop, problem 2

    From a balcony \(14.1\ \text{m}\) above the street a body is thrown vertically upward at \(10\ \text{m/s}\). Calculate the time it takes to reach the ground.

  3. From the roof of a building

    Course workshop, problem 3

    A body is thrown vertically upward at \(98\ \text{m/s}\) from the roof of a building \(100\ \text{m}\) high. Determine (a) the maximum height it reaches above the ground, (b) the time when it passes the launch point, (c) the velocity on reaching the ground and (d) the total time until it reaches the ground.

  4. A stone from a tower

    Course workshop, problem 4

    From the top of a tower a stone is thrown up at \(15\ \text{m/s}\); it rises and falls along the outside of the tower. Taking the launch point as origin and \(g = 10\ \text{m/s}^2\), find (a) its position and velocity after 1 s and 4 s, (b) its velocity when it is 8 m above the starting point and (c) the time until it passes that point again. (Choose g = 10 above and query the other instants.)

  5. A water balloon thrown down

    Class notes

    A student throws a water balloon vertically downward from a building with an initial speed of \(6\ \text{m/s}\). Air resistance can be ignored. (a) What is its speed after falling for 2 s? (b) How far does it fall in that time? (c) What is its speed after falling 10 m? (Query the position y = −10 m for part c.)

  6. Keys to a window

    Serway 2.39

    A student throws a set of keys vertically upward to her sorority sister in a window \(4.00\ \text{m}\) above. The keys are caught \(1.50\ \text{s}\) later. (a) With what initial velocity were they thrown? (b) What was their velocity just before they were caught?

  7. Attacking a castle

    Serway 2.42

    An attacker at the base of a castle wall \(3.65\ \text{m}\) high throws a rock straight up with speed \(7.40\ \text{m/s}\) from a height of \(1.55\ \text{m}\) above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is its speed at the top? (The rock must rise \(\Delta y = 2.10\ \text{m}\).)

  8. From the edge of a cliff

    Serway 2.38

    A ball is thrown directly upward with an initial speed of \(8.00\ \text{m/s}\) from a height of \(30.0\ \text{m}\). After what time interval does it strike the ground?

Theory after R. Resnick, D. Halliday and K. Krane, Física, vol. 1, chapter 2 (section 2-7); problems from R. Serway and J. Jewett, Física para ciencias e ingeniería, vol. 1, chapter 2, and from the course notes and kinematics workshop.