Questions:
1. If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x0x0?
2. If you are given values for Δx, v⃗ , and ΔtΔx, v→, and Δt, which kinematic equation could be used to find v⃗ 0v→0?
3. For horizontally-launched projectiles, which of the following describes acceleration in both directions with a⃗ x=0a→x=0 and a⃗ y=−9.8 m/s2a→y=−9.8 m/s2.
4. Your friend says you can’t use the equation Δx=v⃗ 0Δt + 12a⃗ (Δt)2Δx=v→0Δt + 12a→(Δt)2 to find the horizontal displacement of a horizontal projectile with a constant velocity because you don’t know the acceleration in the horizontal direction. Is your friend correct? Why or why not?
Answers:
1.×0=0
2.Δx =(v⃗ +v⃗ 02)Δt
3. Acceleration is constant in both directions.
4. My friend is incorrect. If the velocity is constant, then the horizontal acceleration is zero.
(You can trust my answers if YOU want too)
— Coconut🥥
To explain how to get the answers:
1. The value of x0x0 represents the initial horizontal position of the object. Since the object is projected horizontally, it starts at the same horizontal position as the point of projection. Therefore, the value of x0x0 is 0.
2. To find v⃗0v→0, the initial velocity, we can use the equation Δx=(v⃗+v⃗02)Δt, where Δx is the displacement, v⃗ is the final velocity, v⃗0 is the initial velocity, and Δt is the time interval. Rearranging the equation, we have v⃗0=Δx−(v⃗+v⃗02)Δt.
3. For horizontally-launched projectiles, acceleration in the x-direction (horizontal) is typically considered zero, as there is no force acting on the object horizontally. So, a⃗x=0a→x=0. In the y-direction (vertical), the acceleration is typically considered -9.8 m/s2 due to gravity, so a⃗y=−9.8 m/s2a→y=−9.8 m/s2.
4. Your friend is incorrect. The equation Δx=v⃗0Δt+12a⃗(Δt)2 can indeed be used to find the horizontal displacement of a horizontal projectile with constant velocity. In this case, when the velocity is constant, the horizontal acceleration is zero (a⃗x=0a→x=0), making the second term on the right side of the equation zero. Therefore, the equation simplifies to Δx=v⃗0Δt, which can be used to find the horizontal displacement.
Keep in mind that these explanations provide a general understanding of the concepts involved. In specific scenarios, such as the presence of air resistance, additional factors may need to be considered.