Test your understanding of horizontal projectile motion, from its initial conditions to its flight time, range, velocity components, and parabolic path. Assume air resistance is negligible unless a question states otherwise.
Q1. For a projectile launched horizontally, what is its initial vertical velocity?
A horizontal launch has no vertical component of velocity at release, so ##v_{0y}=0##.
Q2. With the positive y-axis upward, which acceleration components describe a horizontal projectile when air resistance is negligible?
There is no horizontal acceleration, while gravity acts downward. With upward chosen as positive, ##a_y=-g##.
Q3. For a projectile launched horizontally, taking the launch point as the origin and upward as positive, what is the vertical displacement after time ##t##?
For a horizontal launch, the initial vertical velocity is zero. Thus the vertical displacement is ##y=v_{0y}t-\dfrac{1}{2}gt^2=-\dfrac{1}{2}gt^2##.
Q4. For a projectile launched horizontally, what is its vertical velocity after time ##t##, if upward is positive?
For a horizontal launch, ##v_{0y}=0##. Under downward acceleration ##a_y=-g##, the vertical velocity is ##v_y=v_{0y}-gt=-gt##.
Q5. An object is launched horizontally from a height of ##19.6\ \text{m}##. Taking ##g=9.8\ \text{m/s}^2##, how long does it take to reach the ground?
Using ##H=\dfrac{1}{2}gt^2## gives ##t=\sqrt{\dfrac{2H}{g}}=\sqrt{\dfrac{39.2}{9.8}}=2\ \text{s}##.
Q6. A projectile moves horizontally at ##12\ \text{m/s}## for ##2\ \text{s}##. How far does it travel horizontally in that time?
Horizontal speed remains constant, so ##x=v_0t=(12)(2)=24\ \text{m}##.
Q7. A projectile is launched horizontally at ##15\ \text{m/s}## from a height of ##19.6\ \text{m}##. If ##g=9.8\ \text{m/s}^2##, what is its horizontal range?
The flight time is ##t=\sqrt{2H/g}=2\ \text{s}##. Thus the range is ##R=v_0t=(15)(2)=30\ \text{m}##.
Q8. For horizontal launches from the same height, with the same gravitational acceleration, what happens to the horizontal range if the initial horizontal speed is doubled?
For horizontal launches from the same height under the same gravitational acceleration, the time of flight is unchanged. Since ##R=v_0t##, doubling ##v_0## doubles the range.
Q9. If the launch height increases while the horizontal speed remains constant, what happens to the time of flight?
The flight time is ##t=\sqrt{2H/g}##, so increasing ##H## increases the time before impact.
Q10. Two projectiles are launched horizontally from the same height, but with different horizontal speeds. Which statement is correct, neglecting air resistance?
Their vertical motions have the same initial vertical velocity, height, and gravitational acceleration, so their times of flight are equal.
Q11. For a horizontal launch, which equation gives the vertical coordinate ##y## in terms of horizontal coordinate ##x##, using the launch point as the origin and taking upward as positive?
Since ##x=v_0t##, substitute ##t=x/v_0## into ##y=-\dfrac{1}{2}gt^2## to obtain ##y=-\dfrac{gx^2}{2v_0^2}##.
Q12. A projectile is launched horizontally at ##10\ \text{m/s}##. Using ##g=10\ \text{m/s}^2##, how far below the launch point is it when it has traveled ##4\ \text{m}## horizontally?
From ##y=-\dfrac{gx^2}{2v_0^2}##, ##y=-\dfrac{(10)(16)}{2(100)}=-0.8\ \text{m}##. The projectile is ##0.8\ \text{m}## below the launch point.
Q13. A projectile is launched horizontally at ##8\ \text{m/s}##. While it is still in flight, what are its velocity components after ##1\ \text{s}## if ##g=9.8\ \text{m/s}^2## and upward is positive?
There is no horizontal acceleration, so ##v_x=8\ \text{m/s}##. With upward positive, the vertical velocity after ##1\ \text{s}## is ##v_y=-gt=-(9.8)(1)=-9.8\ \text{m/s}##.
Q14. Which expression gives the projectile's speed after time ##t## for a horizontal launch at speed ##v_0##?
The velocity components are ##v_x=v_0## and ##v_y=-gt##. The speed is their magnitude, ##\sqrt{v_x^2+v_y^2}=\sqrt{v_0^2+g^2t^2}##.
Q15. Which equation describes the horizontal position after time ##t## for a horizontal launch from the origin?
There is no horizontal acceleration, so horizontal motion is uniform and ##x=v_0t##.
Q16. Taking the vertical axis as positive upward, what shape does the trajectory have when vertical position is plotted against horizontal position for an ideal horizontal launch?
The trajectory equation ##y=-\dfrac{gx^2}{2v_0^2}## is quadratic in ##x## and curves downward from the launch point.
Q17. Two projectiles are launched horizontally from the same height, and the second has twice the initial speed of the first. How do their flight times compare?
Flight time is determined by vertical motion and height, not horizontal speed. Both therefore have the same time of flight.
Q18. If the launch height is multiplied by four while ##g## stays constant, how does the time of flight change?
Since ##t=\sqrt{2H/g}##, multiplying ##H## by four multiplies the time by ##\sqrt{4}=2##.
Q19. At the instant of a horizontal launch, which statement best describes the projectile's velocity?
A horizontal launch means the initial velocity is entirely horizontal: ##v_{0x}=v_0## and ##v_{0y}=0##.
Q20. Why does the horizontal velocity remain constant in the ideal horizontal-launch model?
With air resistance neglected, gravity provides only vertical acceleration. Thus ##a_x=0## and the horizontal velocity remains constant.