Finding Velocity given angle and distance.

[odyssey4001]

Homework Statement

The questions is, that if you throw something at exactly 45 degrees above the horizontal and it travels 60 m before hitting the ground what's it velocity?

Homework Equations

Physic's equation, see below

The Attempt at a Solution

Horizontal Component
Vh = cos 45
= 0.707

now t = 60/0.707
=84.86

Vertical Component
Vv = 0.707 - gt

What do I do now? Am I right to this point as well?

 

[LowlyPion]

Welcome to PF.

The first thing to do is start over.

Thrown at 45 degrees means that vertical and horizontal speeds are the same.

So how long is it in the air? Vy/g is the time to go up ... so total time (up and down) is 2*Vy/g

And how fast is it going to go 60 m?

a = 60 = Vx*t = Vx*2*Vy/g

Since Vx = Vy then just solve.

 

[odyssey4001]

Welcome to PF.

The first thing to do is start over.

Thrown at 45 degrees means that vertical and horizontal speeds are the same.

So how long is it in the air? Vy/g is the time to go up ... so total time (up and down) is 2*Vy/g

And how fast is it going to go 60 m?

a = 60 = Vx*t = Vx*2*Vy/g

Since Vx = Vy then just solve.

Ok cool, so is Vy cos 45 = 0.707? so then Vy = Vx = 0.707?

 

[LowlyPion]

Ok cool, so is Vy cos 45 = 0.707? so then Vy = Vx = 0.707?

No.

.707 is merely the value of cos45 and sin45.

Your mission is to figure out what the initial velocity is.

 

[odyssey4001]

No.

.707 is merely the value of cos45 and sin45.

Your mission is to figure out what the initial velocity is.

ok so is this right,

Vo = \sqrt{(g*d)/(2 sin Theta*cos Theta)}

Vo = \sqrt{(9.8*60) / (2 sin 45*cos 45)}

Vo = 24.24

 

[LowlyPion]

That would be correct.