# Simple projectile motion problem

• tuzobarca
In summary, a baseball is hit at a height h = 1.20 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.2 s after it is hit and then down past the top of the wall 4.1 s later, at distance D = 55 m farther along the wall. The ball's initial velocity is v0x = 32.1606 m/s and its final velocity is v0y = -9.8 m/s.

## Homework Statement

In the figure, a baseball is hit at a height h = 1.20 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.2 s after it is hit and then down past the top of the wall 4.1 s later, at distance D = 55 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?
http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c04/fig04_48.gif

## Homework Equations

For part A I'm using the following equations
r=r0+v0+1/2at2
also
v0x=rf-r0/t

## The Attempt at a Solution

I have been working this problem for over an two hours now and even though it seems simple i keep coming up with the wrong answers.
So, first i know that i have to split up the problem into two parts:
x-components
there should be no acceleration in the x direction, so at constant velocity
the vx component should be simply distance traveled/ time.
I use the section that states the distance between 1.2 and 4.1 seconds as 55m so t=4.1-1.2=2.9s → 55m/2.9s=18.97m/s. This is the vx component of the initial velocity vector.
Y-components:
I use this equation: rf=r0+v0t+1/2at2
rearranging for vnot=-1/2at2/t.
I know a=-9.8m/s, t will be the sum of the sections of time so 1.2s + 2.9s + 1.2s = 5.3s
the Δr should be zero, since it will end up falling to the same height that it left from.
Plugging in i got 25.97m/s.
(B) The initial velocity vector should simply be the sqr(v0x2+v0y2) which gives me 32.1606 m/s
(C)To find the angle of the initial velocity vector i take the inverse tangent of the component vector to obtain, 53.85 °
(D) To find the height of the wall i simply add 1.2m to the result from plugging in 1.2 seconds (the point at which it clears the wall) into the y component equation and using the ynot comp vector and solving for Δr which gives me (1.2m+31.164m-7.056m)=25.308m

What am i doing wrong? it seems to me I'm doing every single step right but then again I'm suffering from insomnia so maybe I'm just missing one crucial step and i don't realize it?

I can't see the image.

Last edited:
Ok i figured it out, i was interpreting the wording wrong in the problem statement.
After the projectile clears the height of the wall it comes back down to the wall height again after 4.1 seconds not at4.1s. WOW, i need to get some sleep. Thanks anyway

## What is a simple projectile motion problem?

A simple projectile motion problem is a type of physics problem that involves an object being launched or thrown into the air and following a curved path under the influence of gravity.

## What are the key components of a simple projectile motion problem?

The key components of a simple projectile motion problem are the initial velocity of the object, the angle at which it is launched, and the acceleration due to gravity.

## How is the distance traveled by a projectile calculated?

The distance traveled by a projectile can be calculated using the equation d = v0t + 1/2at2, where d is the distance, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time.

## What is the maximum height reached by a projectile?

The maximum height reached by a projectile is determined by the initial velocity and the launch angle. It can be calculated using the equation h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## How does air resistance affect the motion of a projectile?

Air resistance can affect the motion of a projectile by slowing it down and altering its trajectory. This is because air resistance creates a force in the opposite direction of the projectile's motion, which can cause it to lose speed and change direction.