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?
2s
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.
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.