To calculate the time of flight, horizontal distance, and peak height of the ball, we can use the equations of motion for projectile motion. Here are the steps to solve the problem:
Split the initial velocity into horizontal and vertical components: The horizontal component (Vx) remains constant throughout the motion and is given by: Vx = V * cos(theta) where V is the initial speed (5 m/s) and theta is the angle of projection (30 degrees).
The vertical component (Vy) changes due to the acceleration due to gravity (g) and is given by: Vy = V * sin(theta)
Time of flight (T): The time of flight is the total time the ball is in the air. At the peak of its trajectory, the vertical component of velocity becomes zero. We can use the equation: Vy = V * sin(theta) - g * t Set Vy = 0 to find the time it takes to reach the peak: 0 = V * sin(theta) - g * t_peak t_peak = V * sin(theta) / g
Since the time of flight is symmetrical, the total time of flight is twice the time to reach the peak: T = 2 * t_peak
Horizontal distance (R): The horizontal distance traveled by the ball is given by: R = Vx * T Plug in the values of Vx and T from the previous steps to calculate R.
Peak height (H): The peak height reached by the ball can be determined by finding the vertical displacement during half of the time of flight. We can use the equation: H = Vy^2 / (2 * g) Plug in the value of Vy from step 1 to calculate H.
Now let's calculate the values using the given information:
V = 5 m/s (initial speed) theta = 30 degrees (angle of projection) g = 9.8 m/s^2 (acceleration due to gravity)
Calculations:
Vx = V * cos(theta) = 5 * cos(30) ≈ 4.33 m/s Vy = V * sin(theta) = 5 * sin(30) ≈ 2.5 m/s
t_peak = Vy / g = 2.5 / 9.8 ≈ 0.255 s T = 2 * t_peak ≈ 2 * 0.255 ≈ 0.51 s
R = Vx * T = 4.33 * 0.51 ≈ 2.21 m
H = Vy^2 / (2 * g) = 2.5^2 / (2 * 9.8) ≈ 0.32 m
Therefore, the time of flight is approximately 0.51 seconds, the horizontal distance is approximately 2.21 meters, and the peak height is approximately 0.32 meters.