A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. What is the velocity when the projectile hits the ground? A) V_x = 35.7 m/s, V_y = −22.5 m/s B) V_x = 39.0 m/s, V_y = −22.5 m/s C) V_x = 35.7 m/s, V_y = −39.0 m/s D) V_x = 39.0 m/s, V_y = −35.7 m/s E) V_x = 22.5 m/s, V_y = −22.5 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we can use the equations of projectile motion. We can break the initial velocity into its x and y components:

Vx = V0cosθ = 45.0 m/s cos(30.0) = 39.0 m/s 
Vy = V0sinθ = 45.0 m/s sin(30.0) = 22.5 m/s

Notice that the initial velocity has no horizontal acceleration but a constant vertical acceleration of -9.8 m/s^2 due to gravity. When the projectile hits the ground, the vertical displacement (i.e. the height above ground) is zero. We can use the following kinematic equation to find the time it takes to reach the ground:

Δy = V0yt + ½at^2 
0 = 22.5t - ½(9.8)t^2

Solving for t, we get:

t = 2.29 s

Now that we know the time, we can use the following kinematic equations to find the final velocity:

Vx = V0x = 39.0 m/s (constant horizontal velocity) 
Vy = V0y + at = 22.5 m/s - (9.8 m/s^2)(2.29 s) = - 12.0 m/s (negative velocity means down)

Therefore, the final velocity is:

Vf = sqrt(Vx^2 + Vy^2) = sqrt((39.0 m/s)^2 + (-12.0 m/s)^2) = 40.5 m/s

Choice B is the only option that matches this value.

A Projectile Is Projected from the Origin with a Velocity