Odd projectile motion question.

[letsfailsafe]

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]

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]

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.

 

[letsfailsafe]

So...

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

= 12.544 m

Is it right?

 

[azizlwl]

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]

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.