# Kinematics question: Projectile trajectory given velocity, height at one instant

• edgarpokemon
In summary, the conversation discusses finding the maximum height, horizontal distance, and velocity of a ball that is shot into the air from the ground. Different methods are suggested, including using the equation v^2=vo^2+2a(dy) to find the final vertical component of velocity and considering the entire trajectory from 8.3m to the ground in one calculation. Rounding errors are also mentioned as a potential issue.

#### edgarpokemon

Thread title amended to be more descriptive of the problem

## Homework Statement

A ball is shot from ground into the air. at height 8.3m, its velocity is 7.8i + 6.5j. Find Maximum height rise, horizontal distance traveled and magnitude of velocity just before the ball hits the ground.

## The Attempt at a Solution

So i got the total height. i got 10.45. My question is in part of magnitude of velocity as it hits the ground. So if the ball reaches a height of 10.45m, then would i use v^2=vo^2+2a(dy) to find final vertical component of velocity. So if the ball reaches a maximum height of 10.45m. then isn't dy=10.45? it says in the solution that i should use 8.3m for dy, but why? if the ball is at 10.45?

What is the ball's velocity when it has just returned to 8.3m height?

that would be -6.5m/s. but why would it not work like i want to do it? I noticed that at 10.45, the ball will not have the same speed of 6.5, but greater. so if i calculate the velocity at 10.45 then can i solve for the final velocity of j as the ball hits the ground? problaby not jaja, but why not?

edgarpokemon said:
would i use v^2=vo^2+2a(dy) to find final vertical component of velocity
Yes, that will work, provided you use the right value for v0 there.
However, there is a disadvantage in this method. In declaring the max height to be 10.45 you would have done some rounding. Using that as the starting point for the next stage of calculation unnecessarily introduces some inaccuracy. For that reason, it would be better to start again and consider the whole trajectory from height 8.3m to ground in one calculation. Or use Simon's method, which produces the same equation.

edgarpokemon said:
that would be -6.5m/s. but why would it not work like i want to do it? I noticed that at 10.45, the ball will not have the same speed of 6.5, but greater. so if i calculate the velocity at 10.45 then can i solve for the final velocity of j as the ball hits the ground? problaby not jaja, but why not?
It would totally work but what haruspex said about rounding errors.
At y=10.45m the speed should be less, not more... but both approaches should give the same answers for the speed when the ball hits the ground.
You use the same equation, but vo is different.

There are a number of different approaches. None are more right than the others but some are easier and less prone to error.

aaaa i see i guess i over thinked it too much! i understand now oh my goshhh thanks!