A small mouse is dropped by a hawk who is ascending straight up at constant rate of 1.0 m/s. Neglecting air friction, after 2.0 seconds, what is the velocity of the mouse?
Subtract g*(2 seconds)= 19.6 m/s
from the initial upward velocity
1.0 - 19.6 = ___?
To find the velocity of the mouse after 2.0 seconds, we need to consider two independent motions: the upward motion of the hawk and the downward motion of the mouse.
Since the hawk is ascending at a constant rate, its velocity remains constant throughout. According to the given information, the hawk is ascending at a rate of 1.0 m/s. Therefore, the velocity of the hawk is +1.0 m/s (positive because it is upwards).
The downward motion of the mouse is due to the force of gravity. On the Earth's surface, the acceleration due to gravity is approximately 9.8 m/s², directed downwards. Since the mouse is dropped, it falls freely under gravity with an initial velocity of 0 m/s.
To determine the velocity of the mouse after 2.0 seconds, we can use the formula:
v = u + at
where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the mouse, the initial velocity (u) is 0 m/s, the acceleration (a) is -9.8 m/s² (negative because it is in the opposite direction of the hawk's motion), and the time (t) is 2.0 seconds.
Plugging in the values, we have:
v = 0 + (-9.8) * 2.0
v = -19.6 m/s
Therefore, the velocity of the mouse after 2.0 seconds is -19.6 m/s (negative because it is downward).