---
title: "Ball kicked, find launch speed"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:ball-kicked-speed"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/ball-kicked-speed
site: BCT Engineering
---

# Ball kicked, find launch speed

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2076 Asoj, 4 marks
- Video: [Projectile motion example problem #3: kicking a soccer ball](https://www.youtube.com/watch?v=M8hhKKABZZI) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/ball-kicked-speed
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

The ball at A is kicked such that $\theta_A = 30^\circ$. If it strikes the ground at B having co-ordinates $x = 15$ ft and $y = -9$ ft, determine the speed at which it is kicked. [Figure: ball launched from A at angle $\theta_A$, landing at B which is $x$ to the right and $y$ below A.]

## Short answer

Use the projectile trajectory equation with x = 15 ft, y = -9 ft, θ = 30°, and g = 32.2 ft/s². Solving for initial speed gives 16.5 ft/s.

## Answer

Treat the ball as a projectile. Use ft units with $g=32.2\ \text{ft/s}^2$.

#### Equations of motion (origin at A)
$$
x=(v_A\cos30^\circ)t,\qquad y=(v_A\sin30^\circ)t-\tfrac12 g t^{2}
$$

At B: $x=15$ ft, $y=-9$ ft. Eliminating $t$:

$$
y=x\tan\theta-\frac{g x^{2}}{2v_A^{2}\cos^{2}\theta}
$$

$$
-9=15\tan30^\circ-\frac{32.2(15)^{2}}{2v_A^{2}\cos^{2}30^\circ}
$$

Since $32.2(15)^2=7245$ and $2\cos^2 30^\circ=1.5$:

$$
-9=8.660-\frac{7245}{1.5\,v_A^{2}}=8.660-\frac{4830}{v_A^{2}}\ \Rightarrow\ \frac{4830}{v_A^{2}}=17.660
$$

$$
v_A^{2}=273.5\ \Rightarrow\ v_A=16.54\ \text{ft/s}
$$

Check: $t=\dfrac{15}{16.54\cos30^\circ}=1.047$ s.

**Answer: the ball is kicked at $v_A\approx16.5$ ft/s (time of flight $\approx1.05$ s).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which equations describe the ball with origin at A?
   A. x = v cos(theta) t, y = v sin(theta) t - (1/2) g t^2
   B. x = v sin(theta) t, y = v cos(theta) t - (1/2) g t^2
   C. x = v cos(theta) t, y = v sin(theta) t + (1/2) g t^2
   D. x = (1/2) g t^2, y = v t
   Answer: A. Horizontal motion is uniform; vertical motion decelerates under g. (Key point: Coordinate equations)

2. Which values are substituted for point B?
   A. x = 15 ft, y = -9 ft
   B. x = 15 ft, y = +9 ft
   C. x = 9 ft, y = -15 ft
   D. x = 0, y = -9 ft
   Answer: A. B is 15 ft across and 9 ft below A. (Key point: Landing coordinates)

3. What is the trajectory equation obtained by eliminating t?
   A. y = x tan(theta) - g x^2 / (2 v^2 cos^2(theta))
   B. y = x tan(theta) + g x^2 / (2 v^2 cos^2(theta))
   C. y = x cot(theta) - g x^2 / (2 v^2 sin^2(theta))
   D. y = x sin(theta) - g x^2 / (2 v^2)
   Answer: A. t = x/(v cos theta) substituted into the y-equation gives this form. (Key point: Eliminating time)

4. Which g is used, since distances are in ft?
   A. 9.81 ft/s^2
   B. 32.2 ft/s^2
   C. 9.81 m/s^2
   D. 3.28 ft/s^2
   Answer: B. In feet, g = 32.2 ft/s^2, giving v_A = 16.5 ft/s. (Key point: Unit for g)

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