Projectile Motion Basics

If a projectile is launched with an initial vertical velocity (Voy) of 60 m/s, how long will it take to reach its highest point?

• 5 seconds

• 6 seconds

• 7 seconds

• 7.5 seconds

At which of the following places does a projectile have the greatest acceleration?

• At the peak of its trajectory

• Half way to the peak

• At the instant it is launched

• All points have the same acceleration

Tags

NGSS.HS-PS2-1

The acceleration during the whole trip, on the horizontal component, is always...

• 0 m/s²

• 3.6 m/s²

• 5 m/s²

• -9.8 m/s²

Tags

NGSS.HS-PS2-1

The final velocity on the highest point is...

• 2.1 m/s

• positive

• 0 m/s

• negative

What is the vertical component of a projectile's velocity if it is launched at an angle of 45° with a speed of 20 m/s?

• 10 m/s

• 14.1 m/s

• 20 m/s

• 28.3 m/s

Tags

DOK Level 2: Skill/Concept

Truman launches a ball at an angle of 45° with a speed of 20 m/s. What is the vertical component of the ball's velocity?

• 10 m/s

• 14.1 m/s

• 20 m/s

• 28.3 m/s

Answer explanation

To find the vertical component of the velocity, use the formula: v_y = v * sin(θ). Here, v = 20 m/s and θ = 45°. Thus, v_y = 20 * sin(45°) = 20 * 0.7071 ≈ 14.1 m/s, which is the correct answer.

Tags

DOK Level 2: Skill/Concept

Alan launches a projectile at an angle of 30° with an initial velocity of 50 m/s. What is the horizontal component of its velocity?

• 25 m/s

• 43.3 m/s

• 50 m/s

• 30 m/s

Answer explanation

To find the horizontal component of the velocity, use the formula: Vx = V * cos(θ). Here, V = 50 m/s and θ = 30°. Thus, Vx = 50 * cos(30°) = 50 * 0.866 = 43.3 m/s, which is the correct answer.

Tags

DOK Level 2: Skill/Concept

Which of the following is true about the horizontal motion of a projectile?

• It has a constant velocity.

• It has a constant acceleration.

• It has zero velocity.

• It has increasing velocity.

Tags

DOK Level 1: Recall

• 1.01 s

• 1.43 s

• 2.02 s

• 3.02 s

Tags

DOK Level 2: Skill/Concept

• 1.01 s

• 1.43 s

• 2.02 s

• 3.02 s

Answer explanation

To find the time to hit the ground, use the formula \( t = \sqrt{\frac{2h}{g}} \). Substituting \( h = 10 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \), we get \( t \approx 1.43 \, \text{s} \). Thus, the correct answer is 1.43 s.

Tags

DOK Level 2: Skill/Concept