a stone is thrown horizontally with an initial velocity of 15m/s from a tower 20 meters high. How long does the stone take to reach the ground?
9 years ago
2 years ago
2s
1 year ago
To find the time it takes for the stone to reach the ground, we can use the equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
- y is the final vertical position of the stone (0 in this case, as it reaches the ground)
- y0 is the initial vertical position of the stone (20 meters)
- v0y is the vertical component of the initial velocity of the stone (0 m/s, as it is thrown horizontally)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken for the stone to reach the ground (what we want to find)
Plugging in the values, we have:
0 = 20 + 0 * t - (1/2) * (-9.8) * t^2
Simplifying the equation:
0 = 20 - 4.9 * t^2
Rearranging the equation:
4.9 * t^2 = 20
Dividing both sides by 4.9:
t^2 = 20 / 4.9
t^2 = 4.081632653
Taking the square root of both sides:
t = √(4.081632653)
t ≈ 2.02 seconds
Therefore, the stone takes approximately 2.02 seconds to reach the ground.
1 year ago
To find the time it takes for the stone to reach the ground, we can use the kinematic equation for vertical motion:
h = h0 + v0yt + 1/2gt^2
Where:
- h is the final height (which is 0 since it reaches the ground).
- h0 is the initial height (20 meters).
- v0y is the vertical component of the initial velocity (0 m/s since it is thrown horizontally).
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t is the time it takes for the stone to reach the ground.
Plugging in the values, we get:
0 = 20 + (0)t + 1/2(9.8)t^2
Simplifying the equation, we have:
9.8t^2 = -20
Dividing both sides by 9.8, we get:
t^2 = -20/9.8
Taking the square root of both sides, we find:
t = √(-20/9.8)
Since time cannot be negative in this context, we can discard the negative solution. Hence, time t ≈ √(2.04) ≈ 1.43 seconds.
Therefore, the stone takes approximately 1.43 seconds to reach the ground.