An arrow fired vertically upwards falls to the ground 6 seconds after starting the movement. Determine the starting and ending speed of the boom and the maximum lift height.
t = 6 seconds – the time interval after which the arrow, fired vertically upward, falls to the ground.
It is required to determine v (m / s) – initial speed, v1 (m / s) – final speed and h (meter) – maximum lifting height.
Since the condition of the problem is not specified, we will not take into account air resistance.
Then, the flight time of the arrow to the ground to the maximum height will be equal to:
t1 = t / 2 = 6/3 = 3 seconds.
Then the initial speed will be equal to:
v = g * t, where g = 9.8 m / s2.
v = 9.8 * 3 = 29.4 m / s (meters per second).
According to the law of conservation of energy, the initial speed of the arrow at the moment it starts moving will be equal to the final speed at the moment it hits the ground, that is:
v1 = v = 29.4 m / s.
The maximum lifting height will be equal to:
h = v2 / (2 * g) = 29.42 / (2 * 9.8) = 864.36 / 19.6 = 44.1 meters.
Answer: the initial and final speed of the boom is 29.4 m / s, the maximum lifting height is 44.1 meters.