A pellet gun is fired straight downward from the edge of a cliff that is 15m above the ground. The pellet strikes the ground with a speed of 27 m/s. How long does it take for the pellet to reach the ground with an initial velocity of 5.9 m/s.
average velocity=(27+5,9)/3=10.97m/s
time=distance/velocity
To solve this problem, we can use the equations of motion in the vertical direction.
The equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and time (t) is given by:
vf = vi + at
In this case, the initial velocity (vi) is 5.9 m/s, the final velocity (vf) is 27 m/s, and the acceleration (a) is due to gravity, which is approximately -9.8 m/s^2 (negative sign indicates downward direction).
So, applying the equation:
27 = 5.9 + (-9.8)t
To find the time (t), we need to solve the equation for t.
Subtracting 5.9 from both sides of the equation:
21.1 = -9.8t
Dividing both sides by -9.8:
t = 21.1 / -9.8
Calculating the value:
t ≈ -2.153 seconds.
However, since time cannot be negative in this context, we can exclude this solution.
Therefore, the pellet takes approximately 2.153 seconds to reach the ground.