Question

3.4 projectile motion

1. a projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0° above the horizontal. it strikes a target above the ground 3.00 seconds later. what are the x and y distances from where the projectile was launched to where it lands?

Explanation:

Step1: Find the horizontal (x) component of initial velocity

The initial horizontal velocity \( v_{0x} = v_0 \cos\theta \), where \( v_0 = 50.0 \, \text{m/s} \) and \( \theta = 30.0^\circ \).
\( v_{0x} = 50.0 \times \cos(30.0^\circ) = 50.0 \times \frac{\sqrt{3}}{2} \approx 43.30 \, \text{m/s} \)

Step2: Calculate horizontal distance (x)

In horizontal motion, there is no acceleration (\( a_x = 0 \)), so the distance \( x = v_{0x} \times t \), where \( t = 3.00 \, \text{s} \).
\( x = 43.30 \times 3.00 = 129.9 \, \text{m} \approx 130 \, \text{m} \) (or keep more decimals as needed)

Step3: Find the vertical (y) component of initial velocity

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \)
\( v_{0y} = 50.0 \times \sin(30.0^\circ) = 50.0 \times 0.5 = 25.0 \, \text{m/s} \)

Step4: Calculate vertical distance (y)

Using the kinematic equation \( y = v_{0y}t - \frac{1}{2}gt^2 \), where \( g = 9.8 \, \text{m/s}^2 \)
\( y = 25.0 \times 3.00 - \frac{1}{2} \times 9.8 \times (3.00)^2 \)
\( y = 75.0 - 4.9 \times 9.00 \)
\( y = 75.0 - 44.1 = 30.9 \, \text{m} \)

Answer:

The horizontal distance \( x \approx 130 \, \text{m} \) (or \( 129.9 \, \text{m} \)) and the vertical distance \( y = 30.9 \, \text{m} \).