Projectile Motion: Finding Speed and Direction at 1.41 Seconds After Firing

[atbruick]

Homework Statement

A projectile is fired with an initial speed of 65.8 at an angle of 39.6 above the horizontal on a long flat firing range. Determine the speed and direction (angle) of the projectile 1.41 seconds after firing.

Homework Equations

Vy=Vyinitial-at

The Attempt at a Solution

Before asking these 2 questions, I was asked to find the max height, time spent in air, and the total horizontal distance covered and found correctly 89.8 m, 8.56 sec, and 434 m. At first I plugged in numbers to the equation to get the speed at 1.41 and got an answer of 52 m but that was incorrect. I'm also not sure if I can find the angle at 1.41 seconds without knowing the height.

 

[paco_uk]

The projectile's speed is not the same as Vy. Start from the definition of speed and see what quantities you need to calculate it.

Similarly, what angle are you trying to find? Can you draw a triangle with the relevant quantities?

 

[atbruick]

Well speed is distance traveled/time elapsed. So do we have to use somewhere the equation v^2=vo^2+2a(x-xo) since distance is involved?

 

[paco_uk]

They have given you the time so your original expression for Vy will work fine. The problem is that you also need to take into account the x component of the velocity: Vx.

Your definition of speed is true but I was thinking more about the relationship between speed and velocity (sorry that wasn't very clear). If you can work out the velocity at a given time, do you know how to calculate the speed from that?