Question

A projectile is launched from ground level with an initial speed of 41.5 m/s at an angle of 34.1° above the horizontal. It strikes a target in the air 2.51 s later.

a) What is the horizontal distance from where the projectile was launched to where it hits the target?
b) What is the vertical distance from where the projectile was launched to where it hits the target?

Answer 1

The horizontal distance from the launch point to where the projectile hits the target is approximately 86.59 m, and the vertical distance is approximately 27.83 m.

Step-by-step explanation:

To solve for the horizontal and vertical distances of a projectile, we need to decompose the initial velocity into horizontal and vertical components based on the launch angle. Using trigonometry, these components are found as:

• Vx = V * cos(θ)
• Vy = V * sin(θ)

Given the initial speed (V) of 41.5 m/s and launch angle of 34.1°, we calculate:

• Vx = 41.5 m/s * cos(34.1°) = 34.5 m/s (approx)
• Vy = 41.5 m/s * sin(34.1°) = 23.4 m/s (approx)

To find the horizontal distance, we use the constant velocity in the x-direction:

x = Vx * t = 34.5 m/s * 2.51 s = 86.59 m (approx)

For the vertical distance, we need to account for the acceleration due to gravity, which is -9.8 m/s2 downward. We use the kinematic equation:

y = Vy * t + 0.5 * a * t2 = 23.4 m/s * 2.51 s + 0.5 * (-9.8 m/s2) * (2.51 s)2

y = 58.71 m - 30.88 m = 27.83 m (approx)

a) The horizontal distance from the launch point to where the projectile hits the target is approximately 86.59 m.
b) The vertical distance from the launch point to where the projectile hits the target is approximately 27.83 m.