Projectiles 1 and 2 are launched from level ground at the same time and follow the trajectories shown in the figure. Which one of the projectiles, if either, returns to the ground first and why?

• Projectile 2, because it will have a smaller initial vertical velocity.

• The flight time cannot be compared without knowing the relative initial speeds.

• Projectile 1, because it will have a greater initial speed, which means it will have a larger average speed and therefore spend less time in the air.

• Projectiles 1 and 2 hit the ground at the same time because they have the same initial speed.

Answer explanation

Projectiles 1 and 2 hit the ground at the same time because they have the same initial speed. The time of flight for projectiles is determined by their vertical motion, which is independent of horizontal speed.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4

A ball is projected horizontally with initial speed v from a platform of height h on Earth. The ball's speed immediately before touching the ground is v_e The experiment is repeated on the Moon with the same initial speed and height. The acceleration due to gravity is less on the Moon than on Earth. The ball's speed just before touching the ground is v_m. Which speed is greater and why?

• v_e, because the ball on the Moon travels less distance

• v_m, because the ball on the Moon will travel a greater horizontal distance before reaching the ground

• v_e, because the ball on the Moon achieves a smaller vertical component of velocity before reaching the ground

• v_m, because the ball on the Moon has less acceleration to slow it down when it is on the Moon

Answer explanation

The ball on the Moon, with lower gravity, takes longer to fall, allowing it to travel a greater horizontal distance. This results in a higher speed just before touching the ground, making v_m greater than v_e.

Tags

NGSS.HS-PS2-4

A launcher can launch a projectile with initial speed v₀ from the edge of a tabletop at any angle between 0° and 90° above the horizontal. How does the horizontal component of the projectile's velocity change as a function of launch angle as the angle is increased from 0° to 90°?

• It increases from zero to a maximum value at a constant rate.

• It decreases from a maximum value to zero at a constant rate.

• It increases from zero to a maximum value at a rate with an increasing magnitude.

• It decreases from a maximum value to zero at a rate with an increasing magnitude.

Answer explanation

As the launch angle increases from 0° to 90°, the horizontal component of velocity, given by v₀ * cos θ, decreases from its maximum value at 0° to zero at 90°. Thus, it decreases from a maximum value to zero at a constant rate.

Tags

NGSS.HS-PS2-1

A ball is launched vertically down with an initial speed v₁ from the edge of a table of height h. The ball reaches the ground a time t₁ after it is launched. Which of the following is a correct expression for the instantaneous speed of the ball immediately before it reaches the ground?

• h/t₁

• gt

• v₁ + gt₁

• 0

Answer explanation

The instantaneous speed just before hitting the ground can be found using the equation of motion: final speed = initial speed + acceleration × time. Here, it is v₁ + gt₁, where g is the acceleration due to gravity.

Tags

NGSS.HS-PS2-1

From time t = 0 s to time t = 10 s, an object moves in one dimension on a horizontal surface with a speed that increases as a function of time. Which of the following methods would give the most accurate estimate of the instantaneous acceleration of the object at t = 5s?

• Find the average velocity between t = 5.1 s and t = 4.9 s and divide that value by 0.2 s.

• Find the average velocity between 10 s and t = 0 s and divide the value by 10 s.

• Find the change in velocity between t = 10 s and t = 0 s and divide the value by 10 s.

• Find the change in velocity between t = 5.1 s and t = 4.0 s and divide that value by 0.2 s.

Answer explanation

The most accurate estimate of instantaneous acceleration at t = 5s is obtained by finding the average velocity between t = 5.1 s and t = 4.9 s, as this method closely approximates the derivative of velocity at that point.

Tags

NGSS.HS-PS2-1

A ball is thrown vertically up into the air from a height of 2 m above the ground with an initial velocity of +15 m/s. If air resistance is ignored, which of the following pairs could be a possible height and velocity of the ball at some point along its trajectory after it is thrown but before it reaches the ground?

• height = 12 m; velocity = -5 m/s

• height = 7 m; velocity = -18 m/s

• height = 0 m; velocity = -15 m/s

• height = 2.5 m; velocity = +5 m/s

Answer explanation

At 12 m, the ball is still ascending, and a velocity of -5 m/s indicates it is on its way down. This is possible as the ball can reach a maximum height before descending. Other options either exceed the initial height or have unrealistic velocities.

Tags

NGSS.HS-PS2-4

A golfer strikes a golf ball that is initially on the ground. The ball leaves the ground with initial speed v₀ at an angle θ to the horizontal and travels toward a vertical wall. The ball collides with the wall at the instant it reaches what could have been its highest point if the wall wasn’t there. The ball rebounds from the wall with the same speed it had when it struck the wall, and then lands at the same point from which it was struck. Which of the following is a correct expression for the horizontal distance between the ball's initial position and the wall?

• 
• 
• 
• 

Answer explanation

voy = vo sin theta. when the ball reaches the wall, it is at its highest point so vy = 0. Using the 1st equation in the equation sheet, to solve for t .. then plug in the derived time expression in x = vox t.

Tags

NGSS.HS-PS2-1

Two position vectors, with magnitudes of 5 m and 10 m, point in the directions shown. In figure 1, the vectors point in the same direction and in Figure 2, the vectors point in the opposite directions. If the vectors in each figure are added, how will the magnitude of the vector sum in Figure 1 compare to that of Figure 2?

• Figure 1 = 5 m; Figure 2 = 5 m

• Figure 1 = 15 m; Figure 2 = 15 m

• Figure 1 = 5 m; Figure 2 = 15 m

• Figure 1 = 15 m; Figure 2 = 5 m

Answer explanation

In Figure 1, both vectors add constructively (5 m + 10 m = 15 m). In Figure 2, they add destructively (10 m - 5 m = 5 m). Thus, Figure 1 = 15 m and Figure 2 = 5 m, making the correct choice: Figure 1 = 15 m; Figure 2 = 5 m.

Tags

NGSS.HS-PS2-1

At time t = 0, a projectile is launched from the top of a cliff at an angle of 30 degrees BELOW the horizontal. Which of the following pairs of graphs best represents the horizontal displacement Δx and the vertical velocity component v_vert of the projectile as a function of time t?

• 
• 
• 
• 

Answer explanation

Option A correctly shows that horizontal displacement Δx increases linearly over time, while vertical velocity v_vert increases linearly due to gravity, reflecting the projectile's downward launch at 30 degrees below the horizontal, V_oy > 0.

Tags

NGSS.HS-PS2-4

A projectile is launched from the ground with an initial speed v₀ at an angle θ above the horizontal. If the projectile reaches its maximum height at time t_h, which of the following expressions correctly represents the time t_h, in terms of v₀, θ, and g?

• sin θ Vo/ g

• cos θ Vo/ g

• 2 Vo sin θ / g

• 2 Vo cos θ / g