Worked examples · Sheet 2

Oblique launches from the ground

Theory used: Equations of motion of the projectile Parabolic trajectory Maximum height Time of flight and range Maximum range and complementary angles

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A ball is kicked from the ground with velocity $v_{0} = 20 m/s$ inclined at $θ = 3 0^{\circ}$ to the horizontal. Neglecting air resistance and the size of the ball, compute the time of flight, the maximum height and the range, with landing at the launch height.

Origin at the launch point, $y$ axis pointing upward.

v_{0 x} = v_{0} cos θ = 17.3 m/s, v_{0 y} = v_{0} sin θ = 10.0 m/s

T = \frac{2 v _{0 y}}{g}, H = \frac{v _{0 y}^{2}}{2 g}, R = v_{0 x} T = \frac{v _{0}^{2} sin 2 θ}{g}

T = \frac{2 \cdot 10.0 m/s}{9.81 m/ s ^{2}} = 2.04 s, H = \frac{( 10.0 m/s ) ^{2}}{2 \cdot 9.81 m/ s ^{2}} = 5.10 m

R = \frac{( 20 m/s ) ^{2} sin 6 0 ^{\circ}}{9.81 m/ s ^{2}} = 35.3 m

Trajectory of the ball kicked at

3 0^{\circ}

with

v_{0} = 20 m/s

: maximum height

5.10 m

halfway, range

35.3 m

The ball can be treated as a point and the air does not slow it down, therefore the equations of motion of the projectile hold. The initial velocity is decomposed into its two components; the horizontal component remains constant during the flight, the vertical one determines how high the ball rises and how long it stays in the air. The landing point is at the launch height, so the formulas for the time of flight and range apply, and the maximum height is obtained from the vertical component alone. The ball stays in flight $2.04 s$ , rises to $5.10 m$ and lands at $35.3 m$ . As a check, $v_{0 x} T = 17.3 m/s \cdot 2.04 s = 35.3 m$ , equal to the value given by the formula with $sin 2 θ$ .

Exercise 2

A projectile launched from the ground with velocity of magnitude $v_{0} = 25 m/s$ must land on the ground $R = 50 m$ from the launch point, on level ground. Neglecting air resistance, determine the possible launch angles and the time of flight corresponding to each.

R = \frac{v _{0}^{2} sin 2 θ}{g} ⟹ sin 2 θ = \frac{g R}{v _{0}^{2}}

sin 2 θ = \frac{9.81 m/ s ^{2} \cdot 50 m}{( 25 m/s ) ^{2}} = 0.785

2 θ = 51. 7^{\circ} or 2 θ = 18 0^{\circ} - 51. 7^{\circ} = 128. 3^{\circ}

θ_{1} = 25. 9^{\circ}, θ_{2} = 64. 1^{\circ}

T = \frac{2 v _{0} sin θ}{g}, T_{1} = \frac{2 \cdot 25 \cdot sin 25. 9 ^{\circ}}{9.81} s = 2.22 s

T_{2} = \frac{2 \cdot 25 \cdot sin 64. 1 ^{\circ}}{9.81} s = 4.59 s

The landing point is at the launch height, therefore the range is given by the formula $R = \frac{v _{0}^{2} sin 2 θ}{g}$ , in which the only unknown is the angle. The value $\frac{g R}{v _{0}^{2}} = 0.785$ is less than $1$ , so the target is reachable, and the equation $sin 2 θ = 0.785$ has two solutions with $0^{\circ} < 2 θ < 18 0^{\circ}$ , supplementary to each other. The two launch angles are therefore complementary, $25. 9^{\circ}$ and $64. 1^{\circ}$ , as predicted by the remark on maximum range. The launch at $64. 1^{\circ}$ has a larger vertical component of the velocity, so it stays in flight longer, $4.59 s$ against $2.22 s$ . If $\frac{g R}{v _{0}^{2}} > 1$ the equation would have no solutions, because the maximum range $\frac{v _{0}^{2}}{g} = 63.7 m$ would be less than $R$ .

Exercise 3

A ball is to be kicked from the ground so that it reaches a maximum height $H = 12 m$ and lands $R = 40 m$ from the launch point, at the same height. Neglecting air resistance, compute the magnitude $v_{0}$ of the initial velocity and the launch angle $θ$ .

H = \frac{v _{0 y}^{2}}{2 g} ⟹ v_{0 y} = 2 g H

T = \frac{2 v _{0 y}}{g}, v_{0 x} = \frac{R}{T} = \frac{g R}{2 v _{0 y}}

tan θ = \frac{v _{0 y}}{v _{0 x}} = \frac{2 v _{0 y}^{2}}{g R} = \frac{4 H}{R}, v_{0} = v_{0 x}^{2} + v_{0 y}^{2}

v_{0 y} = 2 \cdot 9.81 \cdot 12 m/s = 15.3 m/s, T = 3.13 s, v_{0 x} = \frac{40 m}{3.13 s} = 12.8 m/s

tan θ = \frac{4 \cdot 12}{40} = 1.2 ⟹ θ = 50. 2^{\circ}, v_{0} = 15. 3^{2} + 12. 8^{2} m/s = 20.0 m/s

The maximum height depends only on the vertical component of the initial velocity, therefore the condition on $H$ determines $v_{0 y}$ . Once $v_{0 y}$ is known, the time of flight is also known, and it equals $3.13 s$ ; in this time the uniform horizontal motion must cover $40 m$ , and from this one obtains $v_{0 x}$ . The ratio of the two components gives the tangent of the angle, and simplifying one finds the relation $tan θ = \frac{4 H}{R}$ , which depends neither on $g$ nor on $v_{0}$ . So the ball must be kicked at $50. 2^{\circ}$ with velocity $20.0 m/s$ . As a check, with these values the range formula gives $\frac{20. 0 ^{2} \cdot sin 100. 4 ^{\circ}}{9.81} m = 40.1 m$ , in agreement with the data within rounding.