Motion In Two Or Three Dimensions

Velocity components: $${v}_{x}=\frac{dx}{dt}=6t$$ $${v}_{y}=\frac{dy}{dt}=4-2t$$ At t = 2 s: $${v}_{x}=6(2)=12\text{ m/s}$$ $${v}_{y}=4-2(2)=0\text{ m/s}$$ Magnitude: $$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{12}^{2}+{0}^{2}}=12.0\text{ m/s}$$ The particle moves entirely in the x-direction at this instant.

The x-component of velocity does not change. Acceleration along y affects only the y-component of velocity. Motion in one dimension is independent of motion in a perpendicular dimension, so aᵧ ≠ 0 has no effect on vₓ.

Starting point: Acceleration is the second derivative of position with respect to time. Step 1 — Differentiate x(t) twice $${v}_{x}=\frac{dx}{dt}=5$$ $${a}_{x}=\frac{d{v}_{x}}{dt}=0$$ Step 2 — Differentiate y(t) twice $${v}_{y}=\frac{dy}{dt}=6{t}^{2}$$ $${a}_{y}=\frac{d{v}_{y}}{dt}=12t$$ Step 3 — Combine components $$a=\sqrt{{a}_{x}^{2}+{a}_{y}^{2}}=\sqrt{0+(12t{)}^{2}}=12t$$ The acceleration magnitude grows linearly with time and points entirely along the y-axis.

The slope of the vy-vs-t graph would double. The slope of a velocity–time graph equals the acceleration, so doubling aᵧ doubles the rate at which vy changes with time.

Both balls reach the ground at the same time. Vertical motion is independent of horizontal motion. Both balls experience the same vertical acceleration (g downward) from the same initial height with zero initial vertical velocity, so both fall for the same duration.

Newton’s second law applies independently along each perpendicular axis. A force in the x-direction produces acceleration only along x. Because no y-component of force is introduced, aᵧ remains unchanged, so vy is unaffected.

Starting point: Kinematic equation relating velocity, initial velocity, acceleration, and displacement. $${v}_{y}^{2}={v}_{0y}^{2}+2{a}_{y}(y-{y}_{0})$$ Step 1 — Set conditions at maximum height At maximum height, vy = 0. Let y₀ = 0 and y = H. Acceleration is aᵧ = −g. $$0=({v}_{0}sin\theta {)}^{2}+2(-g)(H)$$ Step 2 — Solve for H $$2gH={v}_{0}^{2}{sin}^{2}\theta $$ $$H=\frac{{v}_{0}^{2}{sin}^{2}\theta }{2g}$$ Maximum height is directly proportional to v₀². Doubling launch speed quadruples the maximum height.

The time to reach the ground does not change. The time of flight depends only on vertical motion: h = ½gt², so t = √(2h/g). The launch speed is entirely horizontal and does not appear in the vertical equation. Because perpendicular dimensions are independent, changing vₓ cannot affect the vertical fall time.

Yes, it is consistent. When θ = 0, sin 0 = 0, so H = 0. A launch angle of zero means the entire initial velocity is horizontal, so the projectile has no upward velocity component and gains no height. The derived expression correctly predicts this.

Resolve initial velocity: $${v}_{0x}=15cos60^{\circ}=7.50\text{ m/s}$$ $${v}_{0y}=15sin60^{\circ}=13.0\text{ m/s}$$ Horizontal velocity is constant: $${v}_{x}=7.50\text{ m/s}$$ Vertical velocity at t = 1.0 s: $${v}_{y}={v}_{0y}-gt=13.0-10(1.0)=3.00\text{ m/s}$$ Speed: $$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{7.50}^{2}+{3.00}^{2}}=\sqrt{56.25+9.00}=\sqrt{65.25}=8.08\text{ m/s}$$ The speed is 8.08 m/s. This is less than the launch speed because the vertical component has decreased while the horizontal component remains unchanged.