If it takes 10 seconds for the ball to hit the ground, we can calculate the height of the building using the equations of motion.
When the ball is thrown upwards, it will reach its maximum height and then fall back down to the ground. The total time of flight for the ball is the time it takes to reach the maximum height (which we will call t_max) plus the time it takes to fall back down to the ground (which is also t_max).
Since the ball is thrown vertically upwards, its final velocity (v_f) when it reaches the maximum height is 0 m/s. The initial velocity (v_i) is 40 m/s, and the acceleration due to gravity (a) is -9.8 m/s^2 (taking downwards as negative).
Using the equation of motion: v_f = v_i + at, we can find the time taken to reach the maximum height:
0 m/s = 40 m/s + (-9.8 m/s^2) × t_max -40 m/s = -9.8 m/s^2 × t_max t_max = 40 m/s / 9.8 m/s^2 ≈ 4.08 s
The total time of flight is twice the time taken to reach the maximum height:
Total time of flight = 2 × t_max Total time of flight = 2 × 4.08 s ≈ 8.16 s
However, we are given that it takes 10 seconds for the ball to hit the ground. This indicates that the ball takes an additional 10 - 8.16 = 1.84 seconds to fall back down from the maximum height to the ground.
Now, we can calculate the height of the building using the equation of motion: s = v_i × t + (1/2) × a × t^2.
Using t = 1.84 s, v_i = 0 m/s, and a = -9.8 m/s^2:
s = 0 m/s × 1.84 s + (1/2) × (-9.8 m/s^2) × (1.84 s)^2 s ≈ -7.17 m
The negative sign indicates that the height is below the reference point (e.g., the ground level). To get the magnitude of the height, we take the absolute value:
|s| ≈ 7.17 m
Therefore, the approximate height of the building is 7.17 meters.