Baseball Trajectory: Will It Clear the Fence?

A baseball is launched at speed $v_0$ and angle $\theta$ from ground level. Will it clear a fence of height $h_f$ at horizontal distance $d$? This is the classic "home run" question — and it's solved in two short steps using the trajectory equation.

The Physics

Eliminating time from the parametric equations gives the trajectory equation — height as a function of horizontal position:

$y(x) = x\tan\theta - \dfrac{g\,x^{2}}{2 v_0^{2}\cos^{2}\theta}$

To check whether the ball clears the fence, simply evaluate $y(d)$ and compare to $h_f$:

• If $y(d) > h_f$ → HOME RUN ✓
• If $y(d) < h_f$ → the fence stops it ✗

Step-by-Step Solution

Given:

• Launch velocity: $v_0 = 40\;\text{m/s}$
• Launch angle: $\theta = 35°$
• Fence distance: $d = 100\;\text{m}$
• Fence height: $h_f = 3\;\text{m}$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: Does the baseball clear the fence?

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Step 1 — Decompose the launch velocity.

$v_{0x} = 40\cos 35° = 40 \times 0.8192 \approx 32.77\;\text{m/s}$

$v_{0y} = 40\sin 35° = 40 \times 0.5736 \approx 22.94\;\text{m/s}$

Step 2 — Find the time at which the ball reaches the fence.

The horizontal velocity is constant, so:

$t_d = \dfrac{d}{v_{0x}} = \dfrac{100}{32.77} \approx 3.052\;\text{s}$

Step 3 — Verify the ball is still in the air at $t_d$.

Total flight time:

$T = \dfrac{2 v_{0y}}{g} = \dfrac{2 \times 22.94}{9.81} \approx 4.677\;\text{s}$

Since $t_d = 3.052\;\text{s} < T = 4.677\;\text{s}$, the ball is indeed still flying when it reaches the fence's $x$-coordinate. ✓

Step 4 — Find the ball's height at the fence.

Plug $t_d$ into $y(t) = v_{0y}\,t - \tfrac{1}{2}\,g\,t^{2}$:

$y(t_d) = (22.94)(3.052) - \tfrac{1}{2}(9.81)(3.052)^{2}$

$= 70.01 - \tfrac{1}{2}(9.81)(9.314)$

$= 70.01 - 45.69 \approx 24.32\;\text{m}$

Step 5 — Compare to the fence height.

$\Delta = y(t_d) - h_f = 24.32 - 3.00 \approx 21.32\;\text{m}$

The ball is 21.3 m above the top of the fence as it crosses — that's like clearing a fence with a building's worth of room to spare. 💥

Step 6 — (Bonus) How much margin do we have? Find the minimum speed.

Setting $y(d) = h_f$ and solving for the minimum $v_0$ at the same $\theta$:

$h_f = d\tan\theta - \dfrac{g\,d^{2}}{2 v_{\min}^{2}\cos^{2}\theta}$

$\dfrac{g\,d^{2}}{2 v_{\min}^{2}\cos^{2}\theta} = d\tan\theta - h_f$

$v_{\min}^{2} = \dfrac{g\,d^{2}}{2\cos^{2}\theta\,(d\tan\theta - h_f)}$

$v_{\min}^{2} = \dfrac{(9.81)(10000)}{2(0.8192)^{2}(100 \times 0.7002 - 3)} = \dfrac{98100}{2 \times 0.6710 \times 67.02} = \dfrac{98100}{89.94} \approx 1090.7$

$v_{\min} \approx 33.03\;\text{m/s}$

So you'd need at least 33 m/s at this angle to clear the fence — and we have a comfortable 7 m/s of margin.

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Answer: HOME RUN! ✓ At $v_0 = 40\;\text{m/s}$ and $\theta = 35°$, the baseball is at $\approx 24.32\;\text{m}$ above the ground when it reaches the 100 m mark — a margin of 21.3 m above the 3 m fence. The minimum speed needed to clear this fence at this angle is about 33 m/s, so we have plenty of margin.

Try It

• Drop the velocity to about 18 m/s — watch the ball just barely fail to clear the fence (the indicator turns red).
• Raise the fence to 20 m — turns easy home runs into challenging shots.
• Increase the fence distance to 150 m — the fence is now beyond our range and the ball doesn't even reach it.
• The HUD shows the Δ at fence in real time.