Projectile Motion - Is my work correct?

AI Thread Summary
A ball dropped from a height of 4 meters takes approximately 0.89 seconds to hit the ground when using the correct formula for accelerated motion, which is d = Vi*t + ½a*t². The initial velocity is zero, and acceleration due to gravity is typically taken as 9.81 m/s², leading to a more accurate time calculation compared to using a simplified value of 10 m/s². The original attempt at solving the problem incorrectly applied the formula for constant velocity instead of accelerated motion. Additionally, a new question about projectile motion was introduced, emphasizing the importance of starting separate threads for different topics. Accurate formula application is crucial for solving physics problems effectively.
AnomalyCoder
Messages
29
Reaction score
0

Homework Statement


You drop a ball (0.5kg( from a height of 4m. How long did the ball take to hit the ground?

Homework Equations


Assume that initially the velocity is 0 for the following equations below.
\DeltaPosition = \Delta Velocity * \Delta Time
Acceleration = \Delta Velocity / \Delta Time

The Attempt at a Solution


Acceleration is -10m/s due to gravity.
\DeltaX = \Delta V * \Delta T
4 = \Delta V * \Delta T
4 = V * (V/10)
4 = V2/10
40 = V2
Root(40) = V

Then just plug in and solve for time.

Root(40)/(10m/s) = \DeltaT
\DeltaT is approximately .63 seconds.

I don't know if this is right.. seems like a short amount of time to be honest.
 
Physics news on Phys.org
I don't follow your calc. Looks like you tried to use the TEX code and it messed up. I can't seem to do TEX, either. I'm a retired high school teacher so I do really basic solutions. We know d and want t, the motion is accelerated at g. So we need an accelerated motion formula with d and t in it, but not Vf because we don't know the final velocity and don't need to find it. We do know that the initial velocity is zero. That leaves good old d = Vi*t + ½a*t². Cross out the Vi*t term because Vi is zero. Solve the equation for t. I get slightly less than 1 second.
 
Using that I got .89 seconds..
I wonder why my solution is incorrect though.
And sorry about the messed up formatting, I didn't realize it was like that..
After all that effort i put into use the TEX code.. oh well.

Why would this not work? All I'm doing is substituting variables.

A = 10m/s due to grav
D = 4

4 = Velocity * Time
Velocity/ Acceleration = Time
Velocity / 10 = Time

4 = Velocity * Velocity/10
40 = Velocity^2
Velocity = rad(40)

Then plug that value into V/A = T
rad(40)/10 = T
T = approximately .63
 
It isn't correct to use "4 = Velocity * Time". The formula d = vt does not apply to accelerated motion. You should have a formula sheet like this:

Motion at constant speed: d = vt
Accelerated motion: d = Vi*t + ½a*t², a = ΔV/Δt, V = Vi + a*t

In any problem, the first thing you figure out is what kind of motion is involved so you know which formula(s) apply.

I got 0.903 s but I used g = 9.81 instead of your g = 10.
 
That makes sense.
Silly me, I forgot about that..
Thank you! I'll be sure to use that list of formulas.
 
Most welcome.
Add to your list of formulas as you go. When you get more exotic formulas, be sure to write the explanation right beside the formula so you remember them together:
F = ma, a = F/m A force causes a mass to accelerate.
Ek = ½mv² Velocity gives a mass kinetic energy
 
Hi
I am doing my engineering degree and i am stuck on a question i came to the conclusion that the textbook must be wrong . lectures are over and test is soon . can you please help?


A projectile is thrown upwards and passes a certain height after 2 seconds.3seconds later it passes the same pont again.calculate the height of the point above the ground

Answer given(49.05m)
 
Hi ZARman, welcome to Physics Forums.

It might have been better to start a new thread for your new question. You might keep that in mind for future reference :smile:

I've looked at your question and I agree with your textbook's answer. If you're not getting the same results you'll have to post the details of your solution attempt so that we can see what the problem is.
 
Back
Top