To solve the problem, we can use the kinematic equations of motion.
- Displacement after 4 seconds: The initial velocity (u) is 30 m/s, the time (t) is 4 seconds, and the acceleration due to gravity (g) is approximately 9.8 m/s² (assuming no air resistance). We need to find the displacement (s) after 4 seconds.
The equation for displacement is given by: s = ut + (1/2)gt²
Substituting the values: s = (30 m/s)(4 s) + (1/2)(9.8 m/s²)(4 s)² s = 120 m + 78.4 m s ≈ 198.4 m
Therefore, the displacement after 4 seconds is approximately 198.4 meters.
- Velocity after 4 seconds: The initial velocity (u) is 30 m/s, the time (t) is 4 seconds, and the acceleration due to gravity (g) is approximately 9.8 m/s². We need to find the velocity (v) after 4 seconds.
The equation for velocity is given by: v = u + gt
Substituting the values: v = 30 m/s + (9.8 m/s²)(4 s) v = 30 m/s + 39.2 m/s v ≈ 69.2 m/s
Therefore, the velocity after 4 seconds is approximately 69.2 m/s.
- Maximum height attained: To find the maximum height, we need to determine the time it takes for the object to reach the highest point. At this point, the vertical velocity becomes zero.
Using the equation: v = u + gt
At the highest point, v = 0 m/s, and u = 30 m/s (initial velocity).
0 = 30 m/s - 9.8 m/s² * t_max
Solving for t_max: t_max = 30 m/s / 9.8 m/s² t_max ≈ 3.06 s
Now, we can calculate the maximum height (h) using the equation: h = ut_max - (1/2)gt_max²
h = (30 m/s)(3.06 s) - (1/2)(9.8 m/s²)(3.06 s)² h ≈ 46.53 m
Therefore, the maximum height attained by the object is approximately 46.53 meters.
- Time of flight: The time of flight (T) is the total time taken by the object to reach the ground from its initial position.
Since the object was thrown vertically upward, it will take the same amount of time to return to the ground as it took to reach the maximum height.
Therefore, the time of flight (T) is approximately 2 * t_max ≈ 2 * 3.06 s ≈ 6.12 s.
Thus, the time of flight is approximately 6.12 seconds.