Projectile Motion: Angled Launch Quiz
Test your understanding of velocity components, flight time, maximum height, range, and the parabolic trajectory of an angled projectile. Assume air resistance is negligible; range and total flight-time formulas apply when launch and landing elevations are the same.
Q1. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##30^\circ## above the horizontal. What are its initial horizontal and vertical velocity components?
Resolve the launch velocity using ##v_{0x}=v_0\cos\theta## and ##v_{0y}=v_0\sin\theta##. At ##30^\circ##, these are ##20\cos30^\circ=10\sqrt{3}## and ##20\sin30^\circ=10##, respectively.
Q2. A projectile is launched with speed ##12\ \mathrm{m\,s^{-1}}## at ##60^\circ## above the horizontal. What is its initial vertical velocity component?
The vertical component is ##v_{0y}=v_0\sin\theta##. Therefore ##v_{0y}=12\sin60^\circ=12(\sqrt{3}/2)=6\sqrt{3}\ \mathrm{m\,s^{-1}}##, which is option B.
Q3. A projectile has an initial vertical velocity of ##15\ \mathrm{m\,s^{-1}}##. Taking gravitational acceleration as ##10\ \mathrm{m\,s^{-2}}## downward, how long does it take to reach maximum height?
At maximum height the vertical velocity is zero. From ##v_y=v_{0y}-gt##, the time is ##t=v_{0y}/g=15/10=1.5\ \mathrm{s}##.
Q4. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##30^\circ## and lands at its launch elevation. If ##g=10\ \mathrm{m\,s^{-2}}##, what is its total time of flight?
The vertical launch component is ##v_{0y}=20\sin30^\circ=10\ \mathrm{m\,s^{-1}}##. For equal launch and landing elevations, ##T=2v_{0y}/g=2(10)/10=2\ \mathrm{s}##.
Q5. A projectile is launched with vertical velocity component ##10\ \mathrm{m\,s^{-1}}##. What maximum height does it reach if ##g=10\ \mathrm{m\,s^{-2}}##?
At the top, the vertical velocity is zero. Using ##v_y^2=v_{0y}^2-2gH## gives ##H=v_{0y}^2/(2g)=100/20=5\ \mathrm{m}##.
Q6. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##30^\circ## and lands at its launch elevation. With ##g=10\ \mathrm{m\,s^{-2}}##, what horizontal range does it travel?
The range is ##R=v_0^2\sin(2\theta)/g##. Substituting gives ##R=400\sin60^\circ/10=20\sqrt{3}\ \mathrm{m}##.
Q7. For the same launch speed and equal launch and landing elevations, which pair of launch angles gives the same horizontal range?
For a fixed launch speed and equal launch and landing elevations, ##R=v_0^2\sin(2\theta)/g##, so complementary launch angles have the same range. The angles in option B, ##30^\circ## and ##60^\circ##, sum to ##90^\circ##. The revised option C is not a complementary pair.
Q8. For a fixed launch speed on level ground, at what launch angle is the horizontal range greatest?
The range is ##R=v_0^2\sin(2\theta)/g##. It is greatest when ##\sin(2\theta)=1##, which occurs at ##2\theta=90^\circ## and hence ##\theta=45^\circ##.
Q9. Which equation gives the projectile's vertical position ##y## as a function of horizontal position ##x## for a launch from the origin?
Eliminating time between ##x=v_0\cos\theta\,t## and ##y=v_0\sin\theta\,t-\dfrac{1}{2}gt^2## gives ##y=x\tan\theta-\dfrac{gx^2}{2v_0^2\cos^2\theta}##. The negative quadratic term makes the trajectory parabolic and concave downward.
Q10. Neglecting air resistance, what happens to the horizontal component of a projectile's velocity during its flight?
There is no horizontal acceleration in the model described, so the horizontal velocity stays equal to its initial value, ##v_x=v_0\cos\theta##.
Q11. At the highest point of a projectile's path for a nonvertical launch, which statement about its velocity is correct?
At maximum height the vertical velocity momentarily equals zero. For a nonvertical launch, the horizontal component remains constant at ##v_x=v_0\cos\theta## and is nonzero, so the projectile is still moving horizontally.
Q12. A projectile is launched at ##10\ \mathrm{m\,s^{-1}}## at ##30^\circ##. After ##0.5\ \mathrm{s}##, its vertical velocity is what value if ##g=10\ \mathrm{m\,s^{-2}}##?
Initially ##v_{0y}=10\sin30^\circ=5\ \mathrm{m\,s^{-1}}## upward. After ##0.5\ \mathrm{s}##, ##v_y=v_{0y}-gt=5-10(0.5)=0##, so the projectile is at its maximum height.
Q13. For a projectile that returns to its launch elevation, how do the vertical velocities at two points of equal height on the way up and on the way down compare?
The vertical motion is symmetric about the highest point in this ideal model. At equal heights, the vertical velocities have the same magnitude but opposite signs, as follows from ##v_y^2=v_{0y}^2-2gy##.
Q14. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##30^\circ##. What is its horizontal displacement from the launch point after ##2\ \mathrm{s}##, assuming no air resistance?
The horizontal displacement from the launch point is ##x=v_0\cos\theta\,t##. Thus ##x=20\cos30^\circ(2)=20\sqrt{3}\ \mathrm{m}##.
Q15. For the same launch speed and launch elevation, which change is guaranteed to increase the maximum height?
The maximum height is ##H=v_0^2\sin^2\theta/(2g)##. With the same speed, increasing the angle from ##30^\circ## to ##60^\circ## increases ##\sin^2\theta## and therefore increases the maximum height.
Q16. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##30^\circ##. If ##g=10\ \mathrm{m\,s^{-2}}##, what is its maximum height?
The initial vertical velocity is ##v_{0y}=20\sin30^\circ=10\ \mathrm{m\,s^{-1}}##. At maximum height, ##v_y=0##, so ##H=\dfrac{v_{0y}^2}{2g}=\dfrac{10^2}{2(10)}=5\ \mathrm{m}##. The correct option is A: ##5\ \mathrm{m}##.
Q17. A projectile is launched at ##20\ \mathrm{m\,s^{-1}}## at ##60^\circ## and lands at its launch elevation. With ##g=10\ \mathrm{m\,s^{-2}}##, what is its total flight time?
The vertical component is ##v_{0y}=20\sin60^\circ=10\sqrt{3}\ \mathrm{m\,s^{-1}}##. Thus ##T=2v_{0y}/g=2\sqrt{3}\ \mathrm{s}##.
Q18. An object is launched from the origin with speed ##v_0## at angle ##\theta## measured above the horizontal, with gravity acting downward and air resistance neglected. Which pair gives its horizontal and vertical positions after time ##t##?
Resolving the initial velocity for an angle measured above the horizontal gives horizontal and vertical components ##v_0\cos\theta## and ##v_0\sin\theta##. With upward positive and gravity acting downward, ##x=v_0\cos\theta\,t## and ##y=v_0\sin\theta\,t-\dfrac{1}{2}gt^2##. The other options misassign components or omit a required component.
Q19. At a fixed launch speed and with the same launch and landing elevation, what happens to range when the launch angle changes from ##30^\circ## to ##60^\circ##?
The angles are complementary. Since the range depends on ##\sin(2\theta)##, both angles give ##\sin60^\circ## in the range formula, so the ranges are equal.
Q20. A projectile has initial velocity components ##v_{0x}=12\ \mathrm{m\,s^{-1}}## and ##v_{0y}=9\ \mathrm{m\,s^{-1}}##. How long does it take to reach its highest point if ##g=10\ \mathrm{m\,s^{-2}}##?
The time to maximum height depends on the vertical component, not the horizontal one. Setting ##v_y=v_{0y}-gt=0## gives ##t=9/10=0.9\ \mathrm{s}##.