A pebble is dropped from an open window 100 meters above the ground.
a) What is the velocity of the pebble after 2 seconds?
b) What is the velocity of the pebble when it hits the ground?
v = -gt = -9.8*2
100 = 4.9 t^2
t = 4.52 seconds to ground
so
v = -9.8*4.52
To answer these questions, we need to use basic kinematic equations for motion under constant acceleration due to gravity. The acceleration due to gravity near the surface of the Earth is approximately 9.8 m/s², directed downward.
a) To find the velocity of the pebble after 2 seconds, we can use the following kinematic equation:
v = u + a * t
Where:
v is the final velocity
u is the initial velocity (in this case, 0 because the pebble is dropped from rest)
a is the acceleration due to gravity (9.8 m/s²)
t is the time elapsed (2 seconds)
Plugging in the values, we get:
v = 0 + (9.8 m/s²) * (2 s)
v = 0 + 19.6 m/s
v ≈ 19.6 m/s
Therefore, the velocity of the pebble after 2 seconds is approximately 19.6 m/s.
b) To find the velocity of the pebble when it hits the ground, we can use a different kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement:
v² = u² + 2 * a * s
Where:
v is the final velocity
u is the initial velocity (still 0)
a is the acceleration due to gravity (9.8 m/s²)
s is the displacement (100 m, since it falls from a height of 100 meters)
Plugging in the values and solving for v, we get:
v² = 0² + 2 * (9.8 m/s²) * (100 m)
v² = 0 + 1960 m²/s²
v ≈ √1960 m/s
v ≈ 44.27 m/s
Therefore, the velocity of the pebble when it hits the ground is approximately 44.27 m/s.