During projectile motion (ignoring air resistance), the horizontal component of velocity

Stays constant throughout the flight. In the ideal model, gravity acts vertically and horizontal acceleration is zero. Horizontal velocity therefore stays constant. Neither a steady increase nor a steady decrease follows from gravity alone. At the highest point, vertical velocity becomes zero. The horizontal component normally remains nonzero for an angled launch.

At the peak of an angled projectile's ideal flight, which statement is correct?

Vertical velocity is zero, but acceleration remains downward. Vertical velocity reaches zero at the peak. Gravity still gives downward acceleration. The horizontal velocity remains unchanged, so total velocity is normally nonzero. Vertical speed is not greatest there. The acceleration is vertical, rather than horizontal.

A ball is thrown vertically upwards at 25.2 m/s. With g = 9.8 m/s², when does it reach the peak?

2.57 s. At the peak, vertical velocity is zero. Time to peak is u/g = 25.2/9.8 ≈ 2.57 s. The 2.52 s value uses g = 10 instead. About 5.14 s is the same-height round trip. About 0.39 reverses the required ratio.

A projectile returns to its launch height in the ideal model. When is total flight time twice the ascent time?

When the model has constant downward gravity and negligible air resistance. Under the stated equal-height model, ascent and descent times match. Total flight time is twice the ascent time. A horizontal launch from an elevated point uses different landing levels. Air resistance can destroy this symmetry. A lower landing point also requires a different vertical calculation.

Two ideal projectiles are launched at the same speed at 30° and 60°. They land at their launch heights. What is equal?

Horizontal ranges. The angles are complementary, so their sin 2θ values are equal. Their horizontal ranges are therefore equal. The 60° launch has a larger vertical component. It reaches a greater maximum height and stays airborne longer. Equal range does not make the two trajectories identical.

An ideal projectile has initial vertical velocity 10 m/s. With g = 10 m/s², how far above launch does it rise?

5 m. At the peak, vertical velocity is zero. Maximum rise is H = uᵧ²/(2g) = 100/20 = 5 m. The numerical value 1 comes from the time-to-peak calculation, not height. Ten and twenty metres do not satisfy the maximum-height equation for these inputs.

One ball is dropped from 20 m while another is launched horizontally from the same height. With g = 10 m/s², when do they land?

Both take 2 s. Both have zero initial vertical velocity and the same vertical acceleration. Their vertical motions match when air resistance is neglected. From 20 = ½ × 10 × t², the falling time is 2 s. Neither ball lands earlier because of horizontal motion. Four seconds does not satisfy the stated fall equation.

For fixed launch speed, which condition supports the familiar 45° maximum-range result?

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For fixed launch speed, which condition supports the familiar 45° maximum-range result?

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