A projectile is projected from the origin with a velocity of 10.0 m/s at an angle of 30.0 degrees above the horizontal.What is the velocity of the projectile 2.00 seconds later?
A) Vx = 8.7 m/s,Vy = -14.6 m/s
B) Vx = 14.6 m/s,Vy = -5.6 m/s
C) Vx = -8.7 m/s,Vy = -14.6 m/s
D) Vx = -14.6 m/s,Vy = 8.7 m/s
E) Vx = 5.6 m/s,Vy = -14.6 m/s

Question

Quizplus · Accepted Answer

Using the equations of motion, we can find the velocity of the projectile in the x and y directions separately.

In the x direction: 
Vx = V0x = V0cosθ = 10.0m/s * cos(30.0°) = 8.66m/s

In the y direction: 
Vy = V0y + g*t 
where V0y = V0*sinθ = 10.0m/s * sin(30.0°) = 5.00m/s 
and g = -9.81m/s^2 (assuming upward is positive)

So, Vy = 5.00m/s - 9.81m/s^2 * 2.00s = -14.62m/s

Therefore, the velocity of the projectile 2.00 seconds later is: 
Vx = 8.66m/s and Vy = -14.62m/s

The answer is choice A.

A Projectile Is Projected from the Origin with a Velocity