a student standing on the roof of a 50 meter high building kicks a stone at a horizontal speed of 4 meters per second, how much time is required for the stone to reach the level ground below?
d=initial velocity times time + (1/2)at^2
-50=0+ (1/2)(-9.81)t^2
t=3.19 seconds
3.19 seconds
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To find the time required for the stone to reach the ground, we can use the equations of motion.
First, let's list the known values:
- Initial vertical position (y0) = 50 meters (height of the building)
- Initial vertical velocity (v0) = 0 m/s (stone is initially at rest in the vertical direction)
- Acceleration due to gravity (g) = 9.8 m/s^2 (acting downwards)
- Initial horizontal velocity (v0x) = 4 m/s (given)
- Vertical final position (y) = 0 meters (level ground)
We can use the vertical position-time equation:
y = y0 + v0y * t - (1/2) * g * t^2
Since the stone starts at rest in the vertical direction, the initial vertical velocity (v0y) is 0 m/s. Also, at the level ground, the final vertical position (y) is 0 meters. We can simplify the equation to solve for time:
0 = 50 + 0 * t - (1/2) * 9.8 * t^2
0 = 50 - 4.9t^2
Simplifying further, we have a quadratic equation:
4.9t^2 = 50
Now, we can solve for time (t). Dividing both sides of the equation by 4.9:
t^2 = 10.2
Taking the square root of both sides:
t = √10.2
Using a calculator, we find that the square root of 10.2 is approximately 3.19.
Therefore, it takes approximately 3.19 seconds for the stone to reach the level ground below.