rom the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is 7.56 m/s. (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion?
(b) After 0.950 s, how far beneath the cliff top is the pebble?
To solve these questions, we need to use the equations of motion for objects in free fall. Let's address them one by one.
(a) The acceleration of the pebble during the downward motion can be calculated using the equation:
a = g
where "a" is the acceleration and "g" is the acceleration due to gravity. On the surface of the Earth, the average value of acceleration due to gravity is approximately 9.8 m/s² directed downward. Therefore, the magnitude of the acceleration is 9.8 m/s², and the direction is downward (in the negative direction).
(b) To calculate how far beneath the cliff top the pebble is after 0.950 s, we need to use the equation for displacement:
Δy = v_0 * t + (1/2) * a * t²
where Δy is the displacement, v_0 is the initial velocity, t is the time, and a is the acceleration.
Given:
v_0 = 7.56 m/s (since the pebble is fired downward, the initial velocity is negative)
t = 0.950 s
a = 9.8 m/s² (downward)
Plugging in the values, we get:
Δy = (-7.56 m/s) * (0.950 s) + (1/2) * (9.8 m/s²) * (0.950 s)²
Simplifying this equation will give us the answer for the displacement of the pebble beneath the cliff top after 0.950 s.