Kinematic equations for max height

  • #1

Homework Statement


1.JPG

2.JPG

Homework Equations




The Attempt at a Solution



So (i) was easy enough and I got a time of 0.67 seconds.

For (ii), since the horizontal time is 0.67 seconds this means the TOTAL time spent in air is also 0.67 seconds, so to calculate max height we split this time in half (since the ball has to go up to reach max height and then fall down again) and use the kinematic equations to find a value of S (distance)

But they have kept the time, t, as 0.67 seconds and not 0.67/2, why is this?
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
They didn't ask for the maximum height. They asked for the vertical displacement when the ball has traveled 1.5 m horizontally (i.e. at the horizontal distance of the ring).
 
  • Like
Likes ravsterphysics
  • #3
They didn't ask for the maximum height. They asked for the vertical displacement when the ball has traveled 1.5 m horizontally (i.e. at the horizontal distance of the ring).

But max vertical displacement occurs when time spent in air is t/2 (compared to horizontal time) and when v=0, if you take a look at this video you can see the guy did the same thing; he's working out vertical displacement using v=0 which is the same as total air time divided by 2?
 
  • #4
gneill
Mentor
20,925
2,867
But they did not ask for maximum displacement. They asked for the displacement at a particular horizontal distance.
 
  • Like
Likes ravsterphysics
  • #5
951
418

Homework Statement


View attachment 111407
View attachment 111408

Homework Equations




The Attempt at a Solution



So (i) was easy enough and I got a time of 0.67 seconds.

For (ii), since the horizontal time is 0.67 seconds this means the TOTAL time spent in air is also 0.67 seconds, so to calculate max height we split this time in half (since the ball has to go up to reach max height and then fall down again) and use the kinematic equations to find a value of S (distance)

But they have kept the time, t, as 0.67 seconds and not 0.67/2, why is this?

Ignoring for the moment the fact that they didn't ask you for the maximum height, your assumption that the maximum height occurs at t/2 is incorrect. That would only be true if it started and ended at the same height. If you actually did need to find the maximum height you would use the initial vertical speed and the known acceleration of gravity.
 
  • Like
Likes ravsterphysics
  • #6
CWatters
Science Advisor
Homework Helper
Gold Member
10,541
2,308
But they have kept the time, t, as 0.67 seconds and not 0.67/2, why is this?

They keep the time at 0.67 seconds because that's how long it takes to get to a horizontal displacement of 1.5 meters.

Aside: In general it's not necessary to split the motion into two phases (up and down) and apply the equations of motion twice. You can just apply the equations of motion once and answer will come out in the wash. Sometimes (not in this problem) you have to solve a quadratic and in that case you might get two answers (for example sometimes when calculating the time when a ball passes a certain height there are two solutions), in that case you may have to think about which is the required answer.
 
  • Like
Likes ravsterphysics

Related Threads on Kinematic equations for max height

Replies
6
Views
875
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
592
  • Last Post
Replies
5
Views
10K
Replies
2
Views
1K
Replies
1
Views
4K
Replies
14
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
9
Views
5K
Top