Worked examples · Sheet 3

Free falls and vertical launches

Theory used: Law of falling bodies Laws of vertical launch Rise time and maximum height Relation between velocity and displacement Meeting problems

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A stone is dropped from rest from a bridge and hits the water after $2.5 s$ . Neglecting air resistance, compute the height of the bridge above the water and the speed with which the stone hits the water.

The $y$ -axis is oriented downward with the origin at the starting point, so $a = + g$ ; the data are $v_{0} = 0$ and $t = 2.5 s$ .

y (t) = \frac{1}{2} g t^{2}, v (t) = g t

h = \frac{1}{2} g t^{2} = \frac{1}{2} \cdot 9.81 m/ s^{2} \cdot (2.5 s)^{2} = 30.7 m

v = g t = 9.81 m/ s^{2} \cdot 2.5 s = 24.5 m/s

Air resistance is negligible, so by the law of falling bodies the stone moves with constant acceleration $g$ directed downward. With the axis oriented downward the acceleration is positive, and the laws of uniformly accelerated motion, with $v_{0} = 0$ and $y_{0} = 0$ , reduce to $y = \frac{1}{2} g t^{2}$ and $v = g t$ . The bridge is $30.7 m$ high and the stone hits the water at $24.5 m/s$ . As a check, the relation $v = 2 g h = 2 \cdot 9.81 \cdot 30.7 m/s$ gives the same value.

Exercise 2

A ball is thrown vertically upward from the ground with velocity $19.6 m/s$ . Neglecting air resistance and using $g = 9.8 m/ s^{2}$ , compute the maximum height, the instant at which the ball returns to the ground and the instants at which it is at a height of $15 m$ , with the corresponding velocities.

The $y$ -axis is oriented upward with the origin at the ground and $t = 0$ at the instant of the throw; the data are $y_{0} = 0$ , $v_{0} = 19.6 m/s$ , $h = 15 m$ .

y (t) = v_{0} t - \frac{1}{2} g t^{2}, v (t) = v_{0} - g t

t_{s} = \frac{v _{0}}{g} = \frac{19.6 m/s}{9.8 m/ s ^{2}} = 2.0 s, h_{m a x} = \frac{v _{0}^{2}}{2 g} = \frac{( 19.6 m/s ) ^{2}}{2 \cdot 9.8 m/ s ^{2}} = 19.6 m

y (t) = 0 ⟺ t (v_{0} - \frac{1}{2} g t) = 0 ⟹ t_{v} = \frac{2 v _{0}}{g} = 4.0 s

\frac{1}{2} g t^{2} - v_{0} t + h = 0 ⟺ t_{1, 2} = \frac{v _{0} \mp v _{0}^{2} - 2 g h}{g}

v_{0}^{2} - 2 g h = 384.16 - 294 m/s = 9.50 m/s

t_{1} = \frac{19.6 - 9.50}{9.8} s = 1.03 s, t_{2} = \frac{19.6 + 9.50}{9.8} s = 2.97 s

v (t_{1, 2}) = v_{0} - g t_{1, 2} = \pm v_{0}^{2} - 2 g h = \pm 9.50 m/s

y = \dfrac{98}{5} t - 4.9 t^{2}

Height of the ball,

y (t) = 19.6 t - 4.9 t^{2}

: the maximum

19.6 m

is at the instant

2 s

, and the line

y = 15

is crossed at the instants

1.03 s

and

2.97 s

, symmetric with respect to

t = 2 s

The laws of vertical launch are applied with the axis upward, hence with acceleration $- g$ . At the highest point the velocity vanishes, and the formulas for the rise time and the maximum height give $2.0 s$ and $19.6 m$ . The ball returns to the ground when $y (t) = 0$ ; the equation has the solution $t = 0$ , which is the throw, and the solution $t_{v} = 4.0 s$ , twice the rise time, as the symmetry of the motion predicts.

The condition $y (t) = 15 m$ is a quadratic equation with positive discriminant, because $15 m$ is less than the maximum height; therefore the ball is at that height twice, on the way up at $1.03 s$ and on the way down at $2.97 s$ . Substituting into the velocity law gives opposite velocities, $+ 9.50 m/s$ on the way up and $- 9.50 m/s$ on the way down; the same magnitude follows from the relation $v^{2} = v_{0}^{2} - 2 g h$ , which does not contain time.

Exercise 3

From the roof of a building $25 m$ high a ball is thrown vertically upward with velocity $8.0 m/s$ . Neglecting air resistance, compute the maximum height reached above the ground, the instant at which the ball hits the ground and the velocity with which it hits it.

The $y$ -axis is oriented upward with the origin at the ground; the data are $y_{0} = 25 m$ and $v_{0} = 8.0 m/s$ .

y (t) = y_{0} + v_{0} t - \frac{1}{2} g t^{2}, v (t) = v_{0} - g t

h_{m a x} = y_{0} + \frac{v _{0}^{2}}{2 g} = 25 m + \frac{( 8.0 m/s ) ^{2}}{2 \cdot 9.81 m/ s ^{2}} = 28.3 m

v_{f}^{2} = v_{0}^{2} + 2 g y_{0} ⟹ v_{f} = - v_{0}^{2} + 2 g y_{0} = - 64 + 490.5 m/s = - 23.548 m/s \approx - 23.5 m/s

v_{f} = v_{0} - g t_{f} ⟹ t_{f} = \frac{v _{0} - v _{f}}{g} = \frac{8.0 + 23.548}{9.81} s = 3.22 s

The motion is a vertical launch from an initial height $y_{0} > 0$ . The maximum height is the starting height increased by the rise $\frac{v _{0}^{2}}{2 g}$ that the ball climbs, so $28.3 m$ . For the arrival velocity the relation between velocity and displacement is used, with displacement $0 - y_{0} = - y_{0}$ and acceleration $- g$ , which gives $v_{f}^{2} = v_{0}^{2} + 2 g y_{0}$ ; the negative root is chosen because on arrival the ball moves downward. The time of flight is then obtained from the velocity law. The ball hits the ground after $3.22 s$ with a velocity of magnitude $23.5 m/s$ , directed downward. As a check, the positive solution of the equation $y (t) = 0$ , that is, of $4.905 t^{2} - 8.0 t - 25 = 0$ , is $t = 3.22 s$ .

Exercise 4

A stone is dropped from rest from a tower $45 m$ high. At the same instant, from the base of the tower, a second stone is thrown vertically upward with velocity $20 m/s$ . Neglecting air resistance, determine the instant and the height at which the two stones meet and their velocities at that instant.

The $y$ -axis is oriented upward with the origin at the base of the tower; the data are $H = 45 m$ and $v_{0} = 20 m/s$ .

y_{1} (t) = H - \frac{1}{2} g t^{2}, y_{2} (t) = v_{0} t - \frac{1}{2} g t^{2}

y_{1} (t) = y_{2} (t) ⟺ H = v_{0} t ⟺ t = \frac{H}{v _{0}} = \frac{45 m}{20 m/s} = 2.25 s

y = H - \frac{1}{2} g t^{2} = 45 m - \frac{1}{2} \cdot 9.81 m/ s^{2} \cdot (2.25 s)^{2} = 20.2 m

v_{1} = - g t = - 22.1 m/s, v_{2} = v_{0} - g t = 20 m/s - 22.1 m/s = - 2.1 m/s

The two stones have the same acceleration $- g$ and are described in the same reference frame. In the meeting equation the terms $- \frac{1}{2} g t^{2}$ appear on both sides and cancel, so the meeting instant is the same as it would be without gravity, and the initial distance $H$ decreases at the constant relative velocity $v_{0}$ . The stones meet after $2.25 s$ at a height of $20.2 m$ . The solution is acceptable because the height is positive, that is, the meeting takes place before the first stone hits the ground. At that instant both velocities are negative, therefore the second stone is already descending too, in agreement with the fact that its rise time, $\frac{v _{0}}{g} = 2.04 s$ , is less than $2.25 s$ .