A ball is launched at 30 m/s at a 40° angle. At the highest point of its trajectory, what is true about its velocity?
AVelocity is zero — the ball momentarily stops before falling
BVelocity equals 30 m/s in the horizontal direction
CVelocity equals 30 cos(40°) horizontally and is nonzero
DOnly the vertical component remains; horizontal velocity has been lost to gravity
At the apex, vertical velocity is zero (the ball is neither rising nor falling), but horizontal velocity is unchanged throughout the flight — there is no horizontal force to alter it. The horizontal component remains v₀cos(40°) for the entire flight. Option A is a very common misconception: students often assume the ball 'stops' at the top. Option D reverses the physics — gravity only acts vertically, so only the vertical component is reduced.
Question 2 Multiple Choice
A golfer hits two balls with identical speed: one at 30°, one at 60°. Ignoring air resistance and assuming flat ground, how do their horizontal ranges compare?
AThe 60° shot travels farther because it stays in the air longer
BThe 30° shot travels farther because it has more horizontal velocity
CThey travel the same horizontal distance
DThe 45° shot would travel farther than either, but we cannot compare these two without calculating
Launch angles symmetric about 45° produce identical range on flat ground. This follows from the range formula R = (v₀² sin 2θ)/g: sin(2×30°) = sin(60°) = sin(120°) = sin(2×60°). Options A and B identify real trade-offs — the 60° shot stays up longer, the 30° shot has higher horizontal speed — but these effects exactly cancel. This symmetry is a beautiful result from the sin(2θ) structure of the range formula.
Question 3 True / False
At the highest point of its trajectory, a projectile has zero velocity.
TTrue
FFalse
Answer: False
Only the vertical component of velocity is zero at the apex — the projectile has stopped rising and hasn't yet begun to fall. The horizontal component, unchanged throughout the flight (no horizontal force acts), remains v₀cosθ. The total velocity at the apex is v₀cosθ, directed purely horizontally. This is the most common single misconception in projectile motion: confusing 'vertical velocity = 0' with 'total velocity = 0.'
Question 4 True / False
A projectile launched at 30° and one launched at 60° from the same point with the same initial speed land at the same horizontal distance on flat ground.
TTrue
FFalse
Answer: True
The horizontal range formula R = v₀²sin(2θ)/g shows that complementary angles (adding to 90°) give the same sin(2θ) value: sin(60°) = sin(120°). The 30° shot has more horizontal speed but less time aloft; the 60° shot has less horizontal speed but more time aloft. These effects precisely cancel, yielding equal range.
Question 5 Short Answer
Explain why time of flight for a projectile is determined entirely by the vertical component of motion, not by the horizontal speed.
Think about your answer, then reveal below.
Model answer: Gravity acts only vertically, so only the vertical motion controls when the projectile hits the ground. The projectile lands when its vertical position returns to zero, which occurs at a time determined by solving y(t) = v₀sinθ·t − ½gt² = 0. The horizontal speed v₀cosθ determines how far the projectile travels during that time, but has no influence on how long it is in the air. This is the independence of horizontal and vertical motion: they share time as a variable, but neither axis's physics depends on the other.
Students sometimes think a faster horizontal speed 'carries' the projectile further and keeps it in the air longer — but horizontal motion has no vertical component and cannot affect vertical acceleration. The time in the air is a vertical problem only; range is then horizontal speed multiplied by that pre-determined flight time.