Projectile Problem No air resistance(At what velocity was object thrown?)

[1irishman]

Homework Statement

An object is thrown into the air from the ground at an angle of 35deg to the horizontal. If this object is in the air for 9.26 s, at what velocity was it thrown?

Homework Equations

vf^2=vi^2+2ad
d=vit +1/2at^2
d=v/t
a=vf-vi/t

The Attempt at a Solution

x motion
---------
vi=vf because no acceleration
t=9.26 s
a=0

y motion
---------
a= -9.8
t=9.26
d=419.7m
vf=90.7m/s
vi=0

I arrived at vertical distance by the following:
90.7^2 = 0 + 2(-9.8)d
d=419.7
-----------------------
I arrived at final vertical velocity by the following:
-9.8=vf-0/9.26
vf=90.7m/s

I don't know if these are correct values or not, but even if they are...i don't know how they arrived at the final answer of 79.1m/s. Please help?

 

[Doc Al]

d=vit +1/2at^2

This is the equation you want to use to analyze the vertical motion. Use this to solve for vi, which will be the vertical component of the initial velocity. (Then use some trig to find the total velocity, given that component.)

 

[1irishman]

don't we have to know the value for d first though before we can solve for vi on the vertical? Is my value of d wrong above?

 

[Doc Al]

Presumably it starts and ends on the ground, so what is d (vertical position) after 9.26 s?

 

[ideasrule]

What's "d" (vertical displacement) if the object goes up and comes down to the same height?

 

[1irishman]

zero?

 

[1irishman]

so vi on vertical is 45.4m/s?

 

[1irishman]

I have horizontal Rx=45.4/sin35=79.1, so the object was thrown initially at 79.1m/s right?

 

[ideasrule]

Yeah. You could have gotten this by noting that gravity decelerates objects at 9.8 m/s every second and that the ball took 9.26/2 s to reach maximum height, so the object was initially moving at 9.8*9.26/2 m/s.

 

[1irishman]

yeah, but that would be too easy! Just kidding, thank you.

 

[1irishman]

If i had used max height for this problem, could I have solved for vi then as well?