AP Physics 1 Exam Question: Velocity of a Projectile

A projectile is launched from the ground at an angle of 45° above the horizontal with an initial velocity of 20 m/s. The acceleration due to gravity is 9.8 m/s². Calculate the velocity of the projectile at the highest point of its trajectory.

Solution:

To solve this problem, we need to break the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the projectile's motion, while the vertical component changes due to acceleration (gravity in this case).

Let's first find the initial vertical velocity component (Vy) and the initial horizontal velocity component (Vx):

Given:

• Initial velocity (V₀) = 20 m/s
• Launch angle (θ) = 45°

The initial vertical velocity component (Vy) can be found using trigonometry:

Vy = V₀ * sin(θ)

Substituting the given values:

Vy = 20 m/s * sin(45°) = 14.1 m/s (rounded to one decimal place)

The initial horizontal velocity component (Vx) can also be found using trigonometry:

Vx = V₀ * cos(θ)

Substituting the given values:

Vx = 20 m/s * cos(45°) = 14.1 m/s (rounded to one decimal place)

Since the projectile reaches its highest point, the vertical velocity component at that point will be zero.

Using the equation of motion for vertical motion:

Vfy = Vy - g * t

At the highest point, Vfy = 0, so:

0 = Vy - g * t

Rearranging the equation:

t = Vy / g

Substituting the known values:

t = 14.1 m/s / 9.8 m/s² = 1.44 s (rounded to two decimal places)

Since the horizontal velocity component remains constant, the horizontal velocity at the highest point (Vx) will be the initial horizontal velocity component (V₀ * cos(θ)).

Therefore, the velocity of the projectile at the highest point is:

V = Vx = 14.1 m/s.

Hence, the velocity of the projectile at the highest point is 14.1 m/s.