014 (part 1 of 2)
You drop a stone from a position of rest.
How far will it fall vertically in the first 2 s?
The acceleration of gravity is 9.8 m/s
2
.
Answer in units of m
015 (part 2 of 2)
If you instead toss it horizontally at 2.7 m/s,
how far will it fall vertically in the first 2 s?
s(t) = 1/2 at^2
plug in your a and t to get the distance
the stone falls at the same speed, regardless of its horizontal speed.
014 (part 1 of 2):
When you drop a stone from a position of rest, it will accelerate due to gravity. The acceleration of gravity is 9.8 m/s^2, and since the stone is in free fall, it will continue to accelerate at this rate. To find out how far it will fall vertically in the first 2 seconds, we can use the formula:
distance = (1/2) * acceleration * time^2
Plugging in the given values, we get:
distance = (1/2) * 9.8 m/s^2 * (2 s)^2
distance = (1/2) * 9.8 m/s^2 * 4 s^2
distance = 19.6 m
Therefore, the stone will fall vertically a distance of 19.6 meters in the first 2 seconds.
015 (part 2 of 2):
If you toss the stone horizontally at a speed of 2.7 m/s, it will still experience the same vertical acceleration due to gravity. However, the horizontal velocity doesn't affect the vertical distance it will fall.
Therefore, regardless of the horizontal velocity, the stone will still fall vertically a distance of 19.6 meters in the first 2 seconds. Gravity doesn't care if the stone has fancy horizontal moves, it still pulls it vertically with the same acceleration.
To find the distance the stone will fall vertically in the first 2 seconds, we can use the formula for the distance covered by an object under constant acceleration:
\[d = \frac{1}{2}gt^2\]
where:
d = distance
g = acceleration due to gravity (9.8 m/s^2)
t = time (2 s)
Using this formula:
\[d = \frac{1}{2}(9.8)(2)^2 = \frac{1}{2}(9.8)(4) = 19.6\ m\]
Therefore, the stone will fall vertically a distance of 19.6 meters in the first 2 seconds.
Now, let's move on to the second part of the question.
If the stone is tossed horizontally at a velocity of 2.7 m/s, the vertical distance it falls in the first 2 seconds will depend on the acceleration due to gravity acting on it.
Since there is no initial vertical velocity, the only force acting on the stone is the acceleration due to gravity. Therefore, the vertical displacement will be the same as the one we calculated in part 1.
So, the stone will still fall 19.6 meters vertically in the first 2 seconds, regardless of its horizontal velocity.
To find the distance the stone will fall vertically in the first 2 seconds when dropped from rest, we can use the equation of motion:
d = (1/2) * g * t^2
where:
d = distance fallen vertically
g = acceleration due to gravity (9.8 m/s^2)
t = time (2 seconds in this case)
Plugging in the values, we have:
d = (1/2) * 9.8 * (2)^2
d = (1/2) * 9.8 * 4
d = 19.6 meters
Therefore, the stone will fall vertically 19.6 meters in the first 2 seconds when dropped from rest.
Now, if the stone is instead tossed horizontally at 2.7 m/s, the vertical distance it falls in the first 2 seconds will still be determined by the force of gravity, and it will not be affected by the horizontal velocity.
So, the answer to the second question remains the same: the stone will fall vertically 19.6 meters in the first 2 seconds, regardless of the horizontal velocity.