Table of Contents

- 1 What will be the kinetic energy of the projectile when it reaches its highest point?
- 2 Which of these is the correct formula for the time of flight of an angular projectile?
- 3 What happens to the horizontal velocity of a projectile?
- 4 What are the possible unknowns of projectile motion?
- 5 What is the maximum height a projectile can reach?
- 6 What is the equation for vertical motion of a projectile?

## What will be the kinetic energy of the projectile when it reaches its highest point?

Therefore, at the highest or at maximum height, the net velocity is u= vocosθ. So, we can say that at the maximum height, the kinetic energy is minimum as the vertical velocity is zero. Minimum kinetic energy = 12m(v0cosθ)2 . So, the correct answer is “Option D”.

### Which of these is the correct formula for the time of flight of an angular projectile?

The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin .

#### What happens to the horizontal velocity of a projectile?

The numerical information in both the diagram and the table above illustrate identical points – a projectile has a vertical acceleration of 9.8 m/s/s, downward and no horizontal acceleration. This is to say that the vertical velocity changes by 9.8 m/s each second and the horizontal velocity never changes.

**How do you calculate projectile time?**

To define the time of flight equation, we should split the formulas into two cases:

- Launching projectile from the ground (initial height = 0)
- t = 2 * V₀ * sin(α) / g.
- Launching projectile from some height (so initial height > 0)
- t = [V₀ * sin(α) + √((V₀ * sin(α))² + 2 * g * h)] / g.

**What is the formula of time in projectile motion?**

Few Examples of Two – Dimensional Projectiles

Quantity | Value |
---|---|

Equation of path of projectile motion | y = (tan θ0)x – gx2/2(v0cosθ0)2 |

Time of maximum height | tm = v0 sinθ0 /g |

Time of flight | 2tm = 2(v0 sinθ0/g) |

Maximum height of projectile | hm = (v0 sinθ0)2/2g |

## What are the possible unknowns of projectile motion?

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile. Examples of this type of problem are

### What is the maximum height a projectile can reach?

Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance).

#### What is the equation for vertical motion of a projectile?

Equations for the Vertical Motion of a Projectile For the vertical components of motion, the three equations are y = viy•t + 0.5*a y *t2 vfy = viy + a y •t

**What is the best way to analyze projectile motion?**

The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.)