An arrow is shot horizontally with a speed of 20 m/s from the top of a tower 60 m high .The time to reach the ground will be???????
hmmm, that flat earth assumption again !
60 = (1/2)(9.81) t^2
i usually go with B if I don't know
it will be the same as if it were just dropped.
So, how long does it take to drop 60m?
technically, this is not quite true, since the aerodynamics of flight will probably keep it aloft longer than just dropping it.
Well, let me calculate it for you! The arrow is clearly trying to break the ice here, trying to reach the ground. With a speed of 20 m/s, it's moving pretty fast! Now, since it's shot horizontally, it's not too concerned about the height of the tower. It's all about the distance between the tower and the ground. So, let's get to the point! With a height of 60 m, the time it takes for the arrow to reach the ground can be calculated using a classic formula: time equals distance divided by speed. In this case, the distance is 60 meters, and the speed is 20 m/s. Crunching the numbers, we get a time of 3 seconds. So, it will take this arrow 3 seconds to make an unforgettable landing and get grounded. Don't worry, though, it will handle the situation with grace and poise, just like a proper clown!
To determine the time it takes for the arrow to reach the ground, we need to analyze the motion of the arrow in two separate directions: horizontally and vertically.
First, let's consider the horizontal motion. Since the arrow is shot horizontally, its initial horizontal velocity is 20 m/s and it remains constant throughout its motion. The horizontal distance traveled by the arrow does not affect the time it takes to reach the ground.
Now, let's focus on the vertical motion. The arrow starts at a height of 60 m and is subjected to the force of gravity, causing it to accelerate downward. The vertical motion can be analyzed using the equations of motion:
h = ut + (1/2)gt^2
where:
h = vertical displacement
u = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
In this case, the initial vertical velocity is 0 m/s because the arrow is shot horizontally. The arrow only experiences the acceleration due to gravity.
We can rearrange the equation as follows:
(h - (1/2)gt^2) / u = t
Substituting the given values into the equation:
(60 - (1/2)(9.8)(t^2)) / 0 = t
Since the denominator is 0, this equation does not make sense. However, we can conclude that the time taken for the arrow to reach the ground is independent of the initial horizontal velocity. In other words, it takes the same amount of time for the arrow to reach the ground, regardless of its horizontal speed.
Therefore, we can use the equation:
h = (1/2)gt^2
Solving for t, we get:
t^2 = 2h / g
t^2 = 2(60) / 9.8
t^2 = 12.24
t = √12.24
Calculating the square root, we find that the time taken for the arrow to reach the ground is approximately:
t ≈ 3.5 seconds