Maximum Height of a Rocket Launch

When you launch something at a steep angle, you give most of the velocity to the vertical component — and the height it reaches grows quadratically with that vertical speed. This is why model rockets, fireworks, and (real) ballistic missiles all use launch angles close to vertical: they want altitude, not range.

The Physics

Only the vertical velocity matters for peak altitude. The horizontal velocity is along for the ride but plays no role in how high the rocket gets. Starting with $v_{0y} = v_0\sin\theta$ and decelerating under gravity:

$v_y(t) = v_{0y} - g\,t$

The peak is when $v_y = 0$, i.e. at $t_{\text{up}} = v_{0y}/g$. The corresponding height comes from $y(t) = v_{0y}\,t - \tfrac{1}{2}\,g\,t^{2}$ evaluated at $t = t_{\text{up}}$:

$H = \dfrac{v_{0y}^{2}}{2g} = \dfrac{v_0^{2}\sin^{2}\theta}{2g}$

Step-by-Step Solution

Given:

• Launch speed: $v_0 = 50\;\text{m/s}$
• Launch angle: $\theta = 80°$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The maximum altitude $H$, the time to reach it $t_{\text{up}}$, and the corresponding range and flight time.

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Step 1 — Find the vertical component of the launch velocity.

$v_{0y} = v_0\sin\theta = 50\sin 80° = 50 \times 0.9848 \approx 49.240\;\text{m/s}$

(Notice that $\sin 80°$ is very close to 1, so almost all of the 50 m/s is going into the vertical direction.)

Step 2 — Find the time to reach the peak.

Set $v_y(t_{\text{up}}) = 0$:

$0 = v_{0y} - g\,t_{\text{up}}$

$t_{\text{up}} = \dfrac{v_{0y}}{g} = \dfrac{49.240}{9.81} \approx 5.020\;\text{s}$

Step 3 — Compute the peak height.

Method 1 (direct from formula):

$H = \dfrac{v_{0y}^{2}}{2g} = \dfrac{(49.240)^{2}}{19.62} = \dfrac{2424.6}{19.62} \approx 123.58\;\text{m}$

Method 2 (substitute $t_{\text{up}}$ into the kinematic equation):

$H = v_{0y}\,t_{\text{up}} - \tfrac{1}{2}\,g\,t_{\text{up}}^{2}$

$= (49.240)(5.020) - \tfrac{1}{2}(9.81)(5.020)^{2}$

$= 247.18 - \tfrac{1}{2}(9.81)(25.20)$

$= 247.18 - 123.60 \approx 123.58\;\text{m} \;\;\checkmark$

Both methods agree.

Step 4 — (Bonus) Compute total flight time and range.

$T = 2\,t_{\text{up}} = 2 \times 5.020 \approx 10.040\;\text{s}$

$v_{0x} = 50\cos 80° \approx 8.682\;\text{m/s}$

$R = v_{0x}\,T = 8.682 \times 10.040 \approx 87.16\;\text{m}$

Step 5 — Compare to the absolute maximum height.

A pure-vertical launch ($\theta = 90°$) maximizes height:

$H_{\max} = \dfrac{v_0^{2}}{2g} = \dfrac{2500}{19.62} \approx 127.42\;\text{m}$

Our 80° launch gets us $123.58 / 127.42 \approx 96.99\%$ of the way there. Just 4% short of the absolute maximum, while still keeping a tiny bit of horizontal range (87 m) — a reasonable compromise for many applications.

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Answer: The rocket reaches a maximum altitude of $H \approx 123.58\;\text{m}$ (about the height of the Statue of Liberty) at $t_{\text{up}} \approx 5.02\;\text{s}$ after launch. It returns to the ground after a total flight time of $\approx 10.04\;\text{s}$, having traveled a horizontal range of $\approx 87.16\;\text{m}$. A pure-vertical launch would only have reached 4% more altitude (127.4 m) but with zero range.

Try It

• Crank the angle to 90° — the rocket flies straight up to $H \approx 127.4\;\text{m}$ and lands right where it took off.
• Compare with 45° — the rocket only reaches half the height (≈ 63.7 m) even at the same speed. This is why real rockets launch nearly vertical.
• The altitude gauge on the right tracks the current height — watch it rise to its peak and fall back.
• Increase the launch speed to 100 m/s — peak height grows by 4× (since $H \propto v_0^{2}$).