A)You drop a stone from a position of rest.
How far will it fall vertically in the first 3 s?
The acceleration of gravity is 9.8 m/s
Answer in units of m
B)If you instead toss it horizontally at 1.8 m/s,
how far will it fall vertically in the first 3 s?
Damon, your answer is wrong
A) When you drop a stone from a position of rest, it will accelerate downward due to gravity. The acceleration due to gravity is given as 9.8 m/sΒ².
To find out how far the stone will fall vertically in the first 3 seconds, we can use the equation of motion:
s = ut + (1/2)atΒ²
where s is the distance, u is the initial velocity (0 m/s), t is the time (3 s), and a is the acceleration due to gravity (9.8 m/sΒ²).
Plugging in the values:
s = 0(3) + (1/2)(9.8)(3Β²)
s = 0 + (1/2)(9.8)(9)
s = 0 + (1/2)(88.2)
s = 0 + 44.1
s = 44.1 m
Therefore, the stone will fall 44.1 meters vertically in the first 3 seconds.
B) If you toss the stone horizontally at 1.8 m/s, the vertical motion of the stone is influenced by gravity but not the horizontal velocity.
Therefore, the stone will fall vertically the same distance as in part A, which is 44.1 meters, regardless of its horizontal velocity.
To find the distance a stone will fall vertically in the first 3 seconds when dropped from rest, we can use the formula for motion under constant acceleration:
π = 0.5ππ‘Β²
Where:
π is the distance the stone falls vertically
π is the acceleration due to gravity (9.8 m/sΒ²)
π‘ is the time (3 s)
Let's calculate the answers to both parts of the question:
A) When dropped from rest:
π = 0.5 * 9.8 * (3^2)
π = 0.5 * 9.8 * 9
π β 44.1 m
Therefore, the stone will fall approximately 44.1 meters vertically in the first 3 seconds when dropped from rest.
B) When tossed horizontally at 1.8 m/s:
In this case, the horizontal velocity does not affect the vertical distance the stone falls since these two motions are independent. The stone will still fall vertically under the influence of gravity alone.
Therefore, the distance the stone falls vertically in the first 3 seconds will be the same as in part A, which is approximately 44.1 meters.
We already discussed the fact that the vertical motion can be solved separately from the constant horizontal speed motion
THERE IS NO DIFFERENCE between A and B
vertical problem:
a = -g = -9.81 m/s^2
v = Vi + a t = -9.81 t = -9.81*3 = - 39.4 m/s
h = Hi + Vi t - 4.9 t^2
h = Hi + 0 - 4.9* 9 = Hi - 44.1
so
Hi initial height - h height at 3 seconds = distance fallen = 44.1 meters