# Projectile motion problem (can someone check my work?)

• kubaanglin
In summary, the ball is hit at an angle of 30° with respect to the horizontal from a location 130 m away from the wall. The wall is 21 m high. The problem is to find the time it takes for the ball to reach the wall, the initial speed of the ball, and the speed of the ball when it reaches the wall. To solve this problem, an equation can be written for the height of the ball at any time t, and then set equal to the height of the wall at time tf (time of flight). This can be solved for tf and used to find the time it takes the ball to reach the wall, as well as the initial speed of the ball. The speed of the ball
kubaanglin

## Homework Statement

A ball is hit at a height of 1.00 m above ground such that it barely clears a wall that is 21.0 m high. This wall is located 130 m away from the location that the ball was hit (home plate). The ball was hit an angle that is 30° with respect to the horizontal. Find the following quantities:

a) The time it takes the ball to reach the wall.
b) The initial speed of the ball after it is hit.
c) The speed of the ball when it reaches the wall.

See below

## The Attempt at a Solution

Can someone check my work?

Your starting premise is incorrect and this invalidates the rest of your work. You say
## v_y^2=v_{0y}^2+2a\Delta x ##. This is OK except that you should have used ##\Delta y## instead. Most importantly, the next equation is unjustified. You say ##0=v_{0y}^2+2a\Delta x ##. Why is ##v_y = 0##? "Barely clears the wall" doesn't mean that the ball clears the wall when it is at maximum height. Also you replace the acceleration with +9.8 m/s2. It should be negative. That's for starters.

You need to write an equation that gives the height of the ball above ground at any time t, then say that at specific time tf (time of flight) the ball is at height 21 m and solve that equation for tf. At some point you should use that in the same time tf the ball has traveled 130 m horizontally.

mechpeac
Couldn't "Barely clears the wall" mean three different things?

Wouldn't the ball barely clear the wall in all three scenarios depicted above where a, b, or c is equal to 130 m and h is equal to 20 m? Is this problem ambiguous in meaning? What qualifies one of the three scenarios over the other two?

I did mean delta y, but wrote x instead. I used the value for delta y for the calculation.

mechpeac
The three situations you show in the previous post are all possibilities - but not for the specific conditions of this problem. For this problem, only one of those will be true. For the distances specified in this problem, I strongly suspect that this problem will be case (c). But working the problem out will let you know for sure.

I still don't understand how not all of the three situations are possible.

The truth is, this question was on a midterm I just took. When I read the question, I immediately thought that "barely clearing the wall" meant that the maximum height of the projectile was the height of the wall. Do I have any standing to argue that my interpretation of the question is valid?

If the angle of the initial velocity, the vertical distance to the wall, and the horizontal distance to the wall are all specified, then it is true that only one of those cases is possible. Even in your picture, you changed the wall height and horizontal distance to create the three cases - for a given angle, θ, and given initial velocity. In your original problem, you are given a specific wall height, initial velocity angle, and horizontal distance. Try to create the 3 cases for those parameters to see it you can make it work.

## 1. What is projectile motion?

Projectile motion is the motion of an object that is launched or thrown and then moves through the air under the force of gravity alone.

## 2. How is projectile motion different from regular motion?

Projectile motion is different from regular motion because it involves both horizontal and vertical components of motion, whereas regular motion only involves one direction of motion.

## 3. What are the key equations used to solve projectile motion problems?

The key equations used to solve projectile motion problems are the equations for displacement, velocity, and acceleration in the x and y directions. These equations take into account the initial velocity, acceleration due to gravity, and time.

## 4. How do you determine the maximum height and range of a projectile?

The maximum height and range of a projectile can be determined by using the equations for displacement and velocity in the y direction. The maximum height is reached when the vertical velocity is equal to 0, and the range is determined by the horizontal displacement at the point when the projectile returns to its initial height.

## 5. What are some common mistakes to avoid when solving projectile motion problems?

Some common mistakes to avoid when solving projectile motion problems include forgetting to account for the effects of air resistance, using incorrect units, and not breaking down the initial velocity into its horizontal and vertical components. It is also important to double check calculations and use the correct equations for the given scenario.

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