Worked examples · Sheet 1

Horizontal launches from a height

Theory used: Projectile motion Equations of motion of the projectile Horizontal launch Speed as a function of height

Exercise 1

A small ball rolls on the top of a table of height $h = 0.80 m$ and leaves its edge with horizontal velocity $v_{0} = 2.5 m/s$ . Neglecting air resistance, compute the fall time and the horizontal distance from the edge of the table of the point where the ball hits the floor.

Origin on the floor, vertically below the edge; $x$ axis horizontal in the direction of motion, $y$ axis pointing upward.

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

y (t_{c}) = 0 ⟹ t_{c} = \frac{2 h}{g}, d = x (t_{c}) = v_{0} \frac{2 h}{g}

t_{c} = \frac{2 \cdot 0.80 m}{9.81 m/ s ^{2}} = 0.404 s, d = 2.5 m/s \cdot 0.404 s = 1.01 m

Air resistance is negligible, therefore after leaving the table the only acceleration is that of gravity and the equations of motion of the projectile apply with $θ = 0$ . Along $x$ the motion is uniform with velocity $v_{0}$ ; along $y$ the motion is uniformly accelerated and starts with zero vertical velocity, because the initial velocity is horizontal. The ball hits the floor when $y = 0$ , and from this condition one obtains the fall time, which depends only on the height of the table. Substituting $t_{c}$ into the equation of motion for $x$ gives the distance, so the ball lands $1.01 m$ from the edge after $0.404 s$ . As a dimensional check, $\frac{h}{g}$ is measured in $\frac{m}{m/ s ^{2}} = s^{2}$ , therefore its square root is a time.

Exercise 2

A rescue plane flies horizontally at height $h = 125 m$ with velocity $v_{0} = 50 m/s$ and must drop a food package onto a point on the ground. Neglecting air resistance, compute at what horizontal distance from the point the package must be released and with what velocity (magnitude and angle below the horizontal) the package hits the ground.

Origin on the ground below the release point; the package starts with the velocity of the plane, $v_{0} = v_{0} \overset{}{^}$ .

t_{c} = \frac{2 h}{g}, d = v_{0} t_{c}, v_{y} = - g t_{c}, v = v_{0}^{2} + 2 g h, tan β = \frac{∣ v _{y} ∣}{v _{0}}

t_{c} = \frac{2 \cdot 125 m}{9.81 m/ s ^{2}} = 5.05 s, d = 50 m/s \cdot 5.05 s = 252 m

∣ v_{y} ∣ = 9.81 m/ s^{2} \cdot 5.05 s = 49.5 m/s, v = 2500 + 2 \cdot 9.81 \cdot 125 m/s = 70.4 m/s

tan β = \frac{49.5}{50} ⟹ β = 44. 7^{\circ}

At the instant of release the package has the same velocity as the plane, therefore it starts with horizontal velocity $v_{0} = 50 m/s$ , and its motion is a horizontal launch from height $h$ . The fall time is obtained from the condition $y = 0$ and equals $5.05 s$ ; in this time the package moves forward horizontally by $252 m$ , so the release must take place $252 m$ before the vertical through the point. For the landing velocity the horizontal component remains $v_{0}$ , while the vertical component is $g t_{c}$ downward. The magnitude is obtained with the Pythagorean theorem or, with the same result, from the speed as a function of height, $v^{2} = v_{0}^{2} + 2 g h$ . The package arrives at $70.4 m/s$ at an angle of $44. 7^{\circ}$ below the horizontal. Since the plane also continues with horizontal velocity $v_{0}$ , during the fall the package always remains vertically below the plane.

Exercise 3

A stone is thrown horizontally from a cliff of height $h = 20 m$ , and the stone falls into the sea at $d = 18 m$ from the base of the cliff. Neglecting air resistance, compute the launch velocity $v_{0}$ and the velocity (magnitude and angle below the horizontal) with which the stone hits the water.

Origin at sea level at the base of the cliff, $y$ axis pointing upward.

t_{c} = \frac{2 h}{g}, v_{0} = \frac{d}{t _{c}} = d \frac{g}{2 h}

t_{c} = \frac{2 \cdot 20 m}{9.81 m/ s ^{2}} = 2.02 s, v_{0} = \frac{18 m}{2.02 s} = 8.91 m/s

∣ v_{y} ∣ = g t_{c} = 19.8 m/s, v = v_{0}^{2} + v_{y}^{2} = 8.9 1^{2} + 19. 8^{2} m/s = 21.7 m/s

tan β = \frac{∣ v _{y} ∣}{v _{0}} = \frac{19.8}{8.91} ⟹ β = 65. 8^{\circ}

Trajectory of the stone,

y = 20 - \frac{20}{324} x^{2}

in meters: it starts horizontally and lands in the water

18 m

from the base, with velocity inclined

65. 8^{\circ}

below the horizontal

The landing distance is known and the launch velocity is unknown, therefore one starts from the vertical motion, which does not depend on $v_{0}$ . The vertical motion is a fall from rest from $20 m$ , so the fall time is $2.02 s$ . In this time the uniform horizontal motion covers $18 m$ , and therefore $v_{0} = \frac{d}{t _{c}} = 8.91 m/s$ . On arrival the horizontal component of the velocity is still $v_{0}$ , the vertical component is $g t_{c}$ downward, and combining them one obtains the magnitude $21.7 m/s$ and the angle $65. 8^{\circ}$ below the horizontal. The check with the speed as a function of height gives $8.9 1^{2} + 2 \cdot 9.81 \cdot 20 m/s = 21.7 m/s$ , in agreement with the result.