The problem involves determining the initial velocity of an object thrown at a 45-degree angle that travels a horizontal distance of 60 meters before hitting the ground. The discussion centers around the application of projectile motion principles.
Some participants have provided guidance on re-evaluating the approach to the problem, suggesting that the vertical and horizontal speeds are equal at a 45-degree launch angle. There is ongoing exploration of the correct relationships and calculations needed to find the initial velocity.
Participants are discussing the implications of using trigonometric functions and the assumptions related to the angle of projection. The conversation reflects a mix of attempts to clarify the problem setup and the mathematical relationships involved.
LowlyPion said: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 said:Ok cool, so is Vy cos 45 = 0.707? so then Vy = Vx = 0.707?
LowlyPion said:No.
.707 is merely the value of cos45 and sin45.
Your mission is to figure out what the initial velocity is.