To find the speed of the ball after 1 second, we need to break down the initial velocity into its horizontal and vertical components.
Given: Initial velocity (v₀) = 10 m/s Launch angle (θ) = 30 degrees Time (t) = 1 second
The horizontal component of velocity (v_x) remains constant throughout the motion. It can be calculated using the formula:
v_x = v₀ * cos(θ)
Substituting the values:
v_x = 10 m/s * cos(30°) v_x = 10 m/s * √3/2 v_x = 5√3 m/s
The vertical component of velocity (v_y) changes due to the effect of gravity. We can calculate the vertical component using the formula:
v_y = v₀ * sin(θ) - g * t
where g is the acceleration due to gravity, approximately 9.8 m/s².
v_y = 10 m/s * sin(30°) - 9.8 m/s² * 1 s v_y = 10 m/s * 0.5 - 9.8 m/s² v_y = 5 m/s - 9.8 m/s² v_y = -4.8 m/s
Note that the negative sign indicates that the velocity is directed downward.
To find the speed after 1 second, we can combine the horizontal and vertical components using the Pythagorean theorem:
speed = √(v_x² + v_y²)
Substituting the values:
speed = √((5√3 m/s)² + (-4.8 m/s)²) speed = √(75 m²/s² + 23.04 m²/s²) speed = √(98.04 m²/s²) speed ≈ 9.90 m/s
Therefore, the speed of the ball after 1 second is approximately 9.90 m/s.