# Odd projectile motion question.

## Homework Statement

Question: "The time of flight of the lure through the air is 3.4 seconds. (g= 9.8m/s/s)
Calculate the maximum height of the lure in its projectile motion."

## Homework Equations

Not sure if I need to use this:
d = (u)(t) + (1/2)(a)(t^2)

## The Attempt at a Solution

I don't know how this really works...

I'm guess to half the time (3.4/2). Thats the time when the object is at rest for a very short time (initial velocity would be 0). And use that equation on to solve the d.

Borek
Mentor
Once the object is at rest (top of the trajectory) what follows (on the vertical axis) is just a free fall, isn't it?

CAF123
Gold Member
You could start by finding the initial vertical component of velocity and using another kinematic relation to find the height. Or do as Borek suggested.

So...

d = (0)(1.6) + (1/2)(9.8)(1.6)^2

= 12.544 m

Is it right?

The time of flight of the lure through the air is 3.4 seconds means the lure left the earth at t=0 and landed back to earth at 3.4sec.

So vertically the final position is back to where it started.
Y position is a horizontal line parallel to x-axis.

Doc Al
Mentor
So...

d = (0)(1.6) + (1/2)(9.8)(1.6)^2

= 12.544 m

Is it right?
Right idea, but 3.4/2 ≠ 1.6. 