Why is the vertical velocity zero at the top but the speed is not?

At the peak only the vertical component vy is zero. The horizontal component stays 17.3 m/s for the whole flight because no horizontal force acts (ignoring air resistance).

Can I use the range formula R = v^2 sin(2θ)/g?

Yes, but only when the ball lands at the same height it was launched from. Here R = 400 × sin 60° / 9.8 = 35.3 m, the same answer.

Physics 1: Ball Kicked at 20 m/s at 30° Above Horizontal: Time of Flight, Maximum Height, and Range

The Problem

A soccer ball is kicked from level ground with a speed of 20 m/s at an angle of 30° above the horizontal. Ignore air resistance and use $g = 9.8\ \text{m/s}^2$. Find (a) the time the ball is in the air, (b) the maximum height it reaches, and (c) how far away it lands.

This is the standard first projectile problem in AP Physics 1 and college Physics 1. The method below works for every launch-from-ground version of it.

Step 1: Split the Initial Velocity Into Components

$$ v_x = 20\cos 30^\circ = 20(0.8660) = 17.32\ \text{m/s} $$

$$ v_{y0} = 20\sin 30^\circ = 20(0.5) = 10.0\ \text{m/s} $$

The horizontal component never changes. The vertical component decreases by 9.8 m/s every second.

Step 2: Time to Reach the Top

At the peak, $v_y = 0$:

$$ 0 = v_{y0} - g t_{\text{up}} \implies t_{\text{up}} = \frac{10.0}{9.8} = 1.02\ \text{s} $$

Step 3: (a) Total Time of Flight

On level ground the trip down takes as long as the trip up:

$$ T = 2t_{\text{up}} = \frac{2(10.0)}{9.8} = \mathbf{2.04\ s} $$

Step 4: (b) Maximum Height

$$ H = \frac{v_{y0}^2}{2g} = \frac{(10.0)^2}{2(9.8)} = \frac{100}{19.6} = \mathbf{5.10\ m} $$

Step 5: (c) Horizontal Range

$$ R = v_x T = 17.32 \times 2.041 = \mathbf{35.3\ m} $$

Check with the level-ground range formula:

$$ R = \frac{v_0^2 \sin 2\theta}{g} = \frac{400 \sin 60^\circ}{9.8} = \frac{400(0.8660)}{9.8} = 35.3\ \text{m} $$

Results at a Glance

Quantity	Formula	g = 9.8	g = 9.81
Time of flight	2 v₀ sin θ / g	2.04 s	2.04 s
Maximum height	(v₀ sin θ)² / (2g)	5.10 m	5.10 m
Range	v₀² sin 2θ / g	35.3 m	35.3 m
Speed at the peak	v₀ cos θ	17.3 m/s	17.3 m/s

Traps That Cost Points

Mixing sine and cosine. Vertical uses $\sin\theta$ when the angle is measured from the horizontal. If the problem gives the angle from the vertical, swap them.
Calculator in radian mode. $\sin 30$ in radians is −0.988. If your vertical component comes out negative, check the mode.
Using the full 2.04 s for the maximum height. The peak happens at half the flight time, 1.02 s.
Using the range formula off a cliff. $R = v_0^2\sin 2\theta/g$ assumes landing height = launch height. For a cliff or a table, solve $y(t) = 0$ for $t$ with the quadratic formula instead.
Saying the speed is zero at the top. Only $v_y$ is zero; the ball still moves horizontally at 17.3 m/s.

Why 30° and 60° Give the Same Range

Because $\sin 2\theta$ is the same for $2(30^\circ) = 60^\circ$ and $2(60^\circ) = 120^\circ$, a ball kicked at 60° with the same speed also lands 35.3 m away. It just goes higher and stays in the air longer. A common multiple-choice question tests exactly this.

The same parabola shows up on the SAT in feet and seconds: see the h(t) = −16t² + 64t + 80 maximum height problem.