If a rock takes 0.750 s to
hit the ground after being thrown down from a height of 4.80 m, determine the rock’s initial velocity.
h = Vo*t + 0.5g*t^2 = 4.80.
Vo*0.75 + 4.9*0.75^2 = 4.8,
Vo = ?
To determine the rock's initial velocity, we can use the kinematic equation for an object in free fall. This equation is:
h = (1/2) * g * t^2 + v_0 * t
where:
h is the height of the object,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken for the object to fall, and
v_0 is the initial velocity of the object.
In this case, we are given:
h = 4.80 m
t = 0.750 s
g = 9.8 m/s^2
We can rearrange the equation to solve for v_0:
h = (1/2) * g * t^2 + v_0 * t
v_0 * t = h - (1/2) * g * t^2
v_0 = (h - (1/2) * g * t^2) / t
Let's substitute the given values into the equation to find the initial velocity:
v_0 = (4.80 - (1/2) * 9.8 * (0.750)^2) / 0.750
v_0 = (4.80 - (1/2) * 9.8 * 0.5625) / 0.750
v_0 = (4.80 - 2.720625) / 0.750
v_0 = 2.079375 / 0.750
v_0 ≈ 2.7725 m/s
Therefore, the rock's initial velocity is approximately 2.7725 m/s.