CLASSICAL-MECHANICS

Equations of Motion for a Projectile

IKey Formulas

Formula Name Description
ax=0,ay=βˆ’ga_x = 0, \quad a_y = -g Accelerations One downward force gives constant vertical acceleration, none horizontal
x(t)=x0+vx,0 tx(t) = x_0 + v_{x,0}\, t Horizontal position Uniform motion at constant speed
y(t)=y0+vy,0 tβˆ’12gt2y(t) = y_0 + v_{y,0}\, t - \tfrac{1}{2} g t^2 Vertical position Constant downward acceleration
vy(t)=vy,0βˆ’gtv_y(t) = v_{y,0} - g t Vertical velocity Slows on the way up, zero at the peak
t1=v0sin⁑θ0gt_1 = \dfrac{v_0 \sin\theta_0}{g} Time to peak When vy=0v_y = 0
ymax⁑=y0+v02sin⁑2θ02gy_{\max} = y_0 + \dfrac{v_0^2 \sin^2\theta_0}{2g} Maximum height Height at the peak

IIInteractive Visualizations

Visualization 1 β€” Launch speed and angle shape the arc

Only the vertical velocity v0sin⁑θ0v_0 \sin\theta_0 decides how high and how long the stone climbs β€” a steeper angle rises higher at the same speed.

πŸ’‘ Challenge: at a fixed speed, find the angle that gives the longest range β€” it is not the one that flies the highest.

Visualization 2 β€” Why the peak is where vy=0v_y = 0

The height curve goes flat at the exact instant the vertical-velocity curve crosses zero β€” both at t1=v0sin⁑θ0/gβ‰ˆ1.44t_1 = v_0\sin\theta_0/g \approx 1.44 s.

πŸ’‘ The height curve's slope is vyv_y itself, so a flat top and a zero crossing mark the same instant.

Visualization 3 β€” A heavy stone and a light pebble fall together

Dividing βˆ’mg=m ay-mg = m\,a_y by the mass leaves ay=βˆ’ga_y = -g for every object, so both masses trace one identical path.

πŸ’‘ This clean result assumes a constant downward gg and no air resistance.

IIIQuiz Questions

Question 1

A stone is thrown with only gravity acting on it. We choose the positive vertical axis to point upward. What are the horizontal and vertical accelerations?

βœ… Correct! No horizontal force gives a_x = 0, and the single downward force gives a_y = -g.

❌ Not quite. Gravity acts only vertically, so the horizontal acceleration must be zero. The vertical acceleration is -g (negative because up is positive).

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Solution:

Gravity is the only force, and it points straight down. Newton's Second Law in components gives:

βˆ’mg=m ayβ‡’ay=βˆ’g-mg = m\,a_y \quad\Rightarrow\quad a_y = -g 0=m axβ‡’ax=00 = m\,a_x \quad\Rightarrow\quad a_x = 0

The vertical acceleration is βˆ’g-g (negative because we chose "up" as positive, so gravity points the other way), and the horizontal acceleration is zero because nothing pushes the stone sideways.

Question 2

A heavy 10 kg stone and a light 0.5 kg pebble are dropped side by side (no air resistance). True or False: the heavy stone has a larger downward acceleration because it weighs more.

βœ… Correct! Mass cancels in -mg = mΒ·a_y, leaving a_y = -g for every object β€” Galileo's result.

❌ Not quite. The heavier stone feels a larger force, but it also has more inertia. Dividing -mg = m·a_y by m removes the mass entirely, so a_y = -g for both.

Show solution

Solution:

Start from Newton's Second Law in the vertical direction: βˆ’mg=m ay-mg = m\,a_y. Dividing both sides by the mass mm gives

ay=βˆ’g,a_y = -g,

with the mass cancelling completely. The acceleration is the same for every object. The heavier stone does feel a larger force (mgmg is bigger), but it also has more inertia (larger mm) to accelerate, and the two effects cancel exactly. Both objects fall with acceleration gg.

The statement is False.

Question 3

A stone is launched at ΞΈ0=45∘\theta_0 = 45^\circ above the horizontal with an initial speed v0=20v_0 = 20 m/s. Using g=9.8g = 9.8 m/sΒ² and sin⁑45βˆ˜β‰ˆ0.707\sin 45^\circ \approx 0.707, how long does it take to reach its highest point?

(Hint: the peak is where the vertical velocity is zero, t1=v0sin⁑θ0/gt_1 = v_0 \sin\theta_0 / g.)

βœ… Correct! t₁ = vβ‚€ sin ΞΈβ‚€ / g = 14.14 / 9.8 β‰ˆ 1.44 s.

❌ Not quite. Use only the vertical part of the velocity: t₁ = vβ‚€ sin ΞΈβ‚€ / g = (20)(0.707)/9.8 β‰ˆ 1.44 s.

Show solution

Solution:

The stone is at its highest point when its vertical velocity is zero:

vy(t1)=v0sin⁑θ0βˆ’gt1=0β‡’t1=v0sin⁑θ0g.v_y(t_1) = v_0 \sin\theta_0 - g t_1 = 0 \quad\Rightarrow\quad t_1 = \frac{v_0 \sin\theta_0}{g}.

Substitute the numbers:

t1=(20)(0.707)9.8=14.149.8β‰ˆ1.44Β s.t_1 = \frac{(20)(0.707)}{9.8} = \frac{14.14}{9.8} \approx 1.44 \text{ s}.

The answer is 1.44 s. (The value 2.04 s comes from forgetting the sin⁑45∘\sin 45^\circ factor and using v0v_0 directly; 2.88 s is the full time of flight, not the time to the peak.)

Question 4

The same stone (v0=20v_0 = 20 m/s, θ0=45∘\theta_0 = 45^\circ) is released from a height of d=2d = 2 m above the ground. Using ymax⁑=d+v02sin⁑2θ02gy_{\max} = d + \dfrac{v_0^2 \sin^2\theta_0}{2g} with g=9.8g = 9.8 m/s² and sin⁑245∘=0.5\sin^2 45^\circ = 0.5, what is the maximum height above the ground?

βœ… Correct! The rise above release is 10.2 m; adding the 2 m release height gives 12.2 m above the ground.

❌ Not quite. The rise above the release point is vβ‚€Β² sinΒ²ΞΈβ‚€ / (2g) = 200/19.6 β‰ˆ 10.2 m. Don't forget to add the 2 m release height to get the height above the ground.

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Solution:

Substitute the peak time into the vertical position equation; it simplifies to

ymax⁑=d+v02sin⁑2θ02g.y_{\max} = d + \frac{v_0^2 \sin^2\theta_0}{2g}.

Plug in the numbers (v02=400v_0^2 = 400, sin⁑245∘=0.5\sin^2 45^\circ = 0.5):

ymax⁑=2+(400)(0.5)2β‹…9.8=2+20019.6=2+10.2=12.2Β m.y_{\max} = 2 + \frac{(400)(0.5)}{2 \cdot 9.8} = 2 + \frac{200}{19.6} = 2 + 10.2 = 12.2 \text{ m}.

The stone rises 10.2 m above the release point, and adding the 2 m release height gives 12.2 m above the ground. The choice 10.2 m forgets to add the release height dd.

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