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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 8

Lesson 5: Horizontally launched projectiles# Horizontally launched projectile review

Review the key concepts, equations, and skills for analyzing horizontally launched projectiles, including how to solve motion problems in two-dimensions using the kinematic formulas.

## Key terms

Term | Meaning | |
---|---|---|

Range | The maximum horizontal distance a projectile travels. |

## Equations

We don't have any new equations, hooray! The equations are the same kinematic formulas as in one dimension, but we now have one set of variables and formulas for each dimension.

### Simplifying the horizontal equations

For the horizontal direction, a, start subscript, x, end subscript is always zero because gravity does not act in this direction. Thus, the kinematic formulas with a, start subscript, x, end subscript terms simplify to:

## How to solve motion problems in two-dimensions

- List our known and unknown variables.
**Note**: the only common variable between the motions is time t. - Break the motion into horizontal and vertical components parallel to the x- and y-axes. Motion in each dimension is independent of the other.
- Solve for the unknowns in the two separate motions—one horizontal and one vertical. We do this using the same procedure as in 1D motion.

## Common mistakes and misconceptions

**Some students forget that motion in the x- and y-direction are independent.**What happens in the x-direction does not affect the y-direction and vice versa.**Make sure to define the coordinate axes and pay attention to the sign of the acceleration constant g.**If upward is positive and a ball falling down toward the Earth, a, start subscript, y, end subscript is minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction because the acceleration is in the negative direction.

## Learn more

For deeper explanations, see our video on horizontally launched projectiles.

To check your understanding and work toward mastering these concepts, check out the exercise on solving kinematic equations for horizontal projectiles.

## Want to join the conversation?

- a plane is travelling with twice the horizontal velocity that is, with a velocity 230 m/s. if all other factors remain the same, determine the time required for the package to hit the ground.(4 votes)
- it's going be the same time for the package to hit the ground, but the horizontal displacement is twice as before(7 votes)

- how does increasing/decreasing the height of a horizontal projectile impact the time and horizontal displacement?(2 votes)
- If you increase the initial height of the projectile, it will take longer to fall to the ground, so the impact time will be larger. If the projectile is in the air longer, it will have more time to travel horizontally. So, the horizontal displacement will be larger.(4 votes)

- Hi, how can you prove the formula for the range of a projectile motion R=(v^2*sin2θ )/g ??(2 votes)
- This should've included the way that the variables were written and worked out in the "Horizontally launched projectile" video so that people know how to solve equations by writing their known and unknown variables.(1 vote)
- how will this affect the trout population?(1 vote)
- It depends on how tough they are.(1 vote)

- in half projectile why initial vertical component is zero(1 vote)
- The initial vertical velocity is 0 because at 0 seconds you are not moving down at all.(1 vote)

- The mounting system for a reciprocating engine weighting 60kg can be considered to be a vertically mounted spring. the spring stiffness is 110kN/m and a maximum amplitude of the vibration when the engine is running is 14mm. The engine has a mass of 250 kg . assume that the vibration can be moddeled by Simple harmonic motion

1.Determine the angular velocity of the vibration spring

2. determine the maximum linear velocity of the vibration

3.calculate the linear velocity of the engine at 4mm from the maximum amplitude

4. Calculate the linear acceleration of the engine at 4mm from the maximum amplitude and the force produced.(1 vote)