Test your projectile problem-solving skills with questions on separating horizontal and vertical motion, handling different launch and landing heights, and interpreting velocity at impact. Use a consistent coordinate system and check that each result makes physical sense.
Q1. Before calculating a projectile's motion, which set of information should you identify first?
These details define the motion and help identify the required displacements and velocity components before choosing equations.
Q2. Why can horizontal and vertical projectile motion be solved separately?
In ideal projectile motion, horizontal and vertical motions are independent, but the projectile's position in both directions is evaluated at the same elapsed time.
Q3. With upward chosen as positive and the launch point as the origin, what is the vertical displacement when a projectile lands below its launch point?
A landing point below the origin has a negative vertical coordinate under this convention.
Q4. With upward chosen as positive and ##g## denoting the positive magnitude of gravitational acceleration, which equation relates vertical displacement to initial vertical velocity and time?
With upward positive, the vertical acceleration is ##a_y=-g##, where ##g## is the positive magnitude. Substituting into ##\Delta y=v_{0y}t+\dfrac{1}{2}a_yt^2## gives ##\Delta y=v_{0y}t-\dfrac{1}{2}gt^2##.
Q5. A ball is launched horizontally from a height of ##19.6\ \text{m}## above level ground. Using ##g=9.8\ \text{m/s}^2##, how long does it take to reach the ground?
The initial vertical velocity is zero. Using ##-19.6=-\dfrac{1}{2}(9.8)t^2## gives ##t^2=4##, so the positive flight time is ##2.0\ \text{s}##.
Q6. A projectile is launched upward with vertical velocity ##10\ \text{m/s}## and lands ##5\ \text{m}## below its launch point. If ##g=10\ \text{m/s}^2##, what is its flight time?
Use ##-5=10t-5t^2##, which gives ##t^2-2t-1=0##. The positive solution is ##t=1+\sqrt{2}\approx2.41\ \text{s}##; the other root is negative and is not a future flight time.
Q7. A projectile is launched vertically upward at ##20\ \text{m/s}## and returns to its launch height. With ##g=10\ \text{m/s}^2##, when does it return to that height?
At the same height, ##\Delta y=0##. Thus ##0=20t-5t^2##, whose nonzero solution is ##t=4\ \text{s}##.
Q8. A projectile has a constant horizontal velocity of ##12\ \text{m/s}## and remains in flight for ##3\ \text{s}##. What horizontal range does it travel?
Horizontal range is ##\Delta x=v_xt##, so ##\Delta x=(12)(3)=36\ \text{m}##.
Q9. A projectile is launched horizontally at ##12\ \text{m/s}## from a height of ##45\ \text{m}##. If ##g=10\ \text{m/s}^2##, what is its horizontal range?
The vertical fall gives ##-45=-5t^2##, so ##t=3\ \text{s}##. The horizontal range is ##(12)(3)=36\ \text{m}##.
Q10. A projectile has horizontal velocity ##8\ \text{m/s}## and vertical velocity ##-6\ \text{m/s}## just before impact. What is its impact speed?
The impact speed is the magnitude of the velocity vector: ##v=\sqrt{8^2+(-6)^2}=10\ \text{m/s}##.
Q11. Just before impact, a projectile's velocity components are ##v_x=8\ \text{m/s}## and ##v_y=-6\ \text{m/s}##. What is its direction relative to the horizontal?
The angle below the horizontal has magnitude ##\tan^{-1}\left(\dfrac{|{-6}|}{8}\right)\approx37^\circ##. The negative vertical component means the projectile is moving downward.
Q12. A projectile is launched horizontally from a cliff. Which value is its vertical velocity at the instant of launch?
A horizontal launch has no initial vertical component. Gravity then changes the vertical velocity during flight.
Q13. Two identical balls start from the same height at the same instant. One is dropped, and the other is launched horizontally. Ignoring air resistance, which reaches the ground first?
Their vertical motions are identical: both start with zero vertical velocity and accelerate downward equally. Horizontal speed does not change the fall time.
Q14. With horizontal acceleration neglected, how does the horizontal velocity of a projectile change during its flight?
Gravity acts vertically, so the horizontal acceleration is zero in this model and horizontal velocity remains constant.
Q15. A projectile is launched at ##20\ \text{m/s}## at an angle of ##30^\circ## above the horizontal. What is its initial vertical velocity component?
The vertical component is ##v_{0y}=v_0\sin\theta##, so ##v_{0y}=20\sin30^\circ=10\ \text{m/s}##.
Q16. A projectile is launched horizontally at ##15\ \text{m/s}## from a point ##20\ \text{m}## above its landing level. Using ##g=10\ \text{m/s}^2##, what is its speed just before impact?
The vertical impact component satisfies ##v_y^2=2g(20)##, giving ##|v_y|=20\ \text{m/s}##. Combining components gives ##v=\sqrt{15^2+20^2}=25\ \text{m/s}##.
Q17. A ball is thrown upward from a platform. With upward positive, which expression correctly represents its vertical velocity after time ##t##?
Vertical acceleration is ##-g## with upward positive. Applying ##v_y=v_{0y}+a_yt## gives ##v_y=v_{0y}-gt##.
Q18. A projectile lands below its launch point. In the upward-positive convention, which sign should be used for its vertical displacement in the vertical-motion equation?
Vertical displacement is final height minus initial height. Since the landing point is lower, ##\Delta y<0## in this coordinate system.
Q19. A vertical-motion equation for a projectile gives two time roots: ##-0.5\ \text{s}## and ##2.5\ \text{s}##. Which root represents a future landing after launch?
The negative root refers to a time before the chosen launch instant. The positive root, ##2.5\ \text{s}##, is the physically relevant future time.
Q20. A student calculates a negative flight time for a projectile launched at the stated instant. What is the best next step?
A negative future flight time is physically inappropriate for the stated launch. Recheck the displacement sign, acceleration sign, and equation before proceeding.