Speed of projectile after t seconds

[Vanessa23]

Homework Statement

A projectile is fired with an initial speed of 68 m/s at 37 degrees above horizontal on a flat firing range. c) Determine speed of the projectile 2.3 seconds after firing. d) Determine direction of projectile 2.3 s after firing.

Homework Equations

c) Vox=Vx=68m/s
Vy=Voy*sin(theta)-gt
V=sqrt(Vx^2+Vy^2)

d) tan^-1(Vy/Vx)= theta

The Attempt at a Solution

I got that Vx=68m/s
Vy=68*sin(37)-(9.8*2.3)
Vy=18.38

V=sqrt(68^2+18.4^2)
V=70.4
but the velocity is not 70.4 what am I doing wrong??
Thanks for any help!
The second part I get theta= 15.1 degrees but I don't want to submit it because if the first part is wrong then this is probably not right either.

 

[LowlyPion]

Homework Statement

A projectile is fired with an initial speed of 68 m/s at 37 degrees above horizontal on a flat firing range. c) Determine speed of the projectile 2.3 seconds after firing. d) Determine direction of projectile 2.3 s after firing.

Homework Equations

c) Vox=Vx=68m/s
Vy=Voy*sin(theta)-gt
V=sqrt(Vx^2+Vy^2)

d) tan^-1(Vy/Vx)= theta

The Attempt at a Solution

I got that Vx=68m/s
Vy=68*sin(37)-(9.8*2.3)
Vy=18.38

V=sqrt(68^2+18.4^2)
V=70.4
but the velocity is not 70.4 what am I doing wrong??
Thanks for any help!
The second part I get theta= 15.1 degrees but I don't want to submit it because if the first part is wrong then this is probably not right either.

You need to take the horizontal component of velocity in the RSS which is 68*Cos 37, not 68.

Then you should be fine.

You will need to get another calculation for theta.

 

[baba89]

Your calculation for the magnitude of the velocity (V) is correct, but you made a mistake in calculating Vy. The correct equation for Vy is V0y + a*t, where V0y is the initial vertical velocity (in this case, 68*sin(37)) and a is the acceleration due to gravity (-9.8 m/s^2). So, the correct calculation for Vy is:
Vy = 68*sin(37) - 9.8*2.3 = 18.38 m/s

Therefore, the magnitude of the velocity is:
V = sqrt(68^2 + 18.38^2) = 70.4 m/s

For part (d), your calculation for the direction (theta) is correct. The direction of the projectile 2.3 seconds after firing is 15.1 degrees above the horizontal.