Question

acceleration due to gravity questions

A rock is dropped from a height of 7meters

a) what is the average velocity of the rock during its fall

b) how long does it take the rock to fall

Answer 1

To solve these problems, we need to use the following kinematic equations:

1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as

where v is final velocity, u is initial velocity, a is acceleration, t is time, and s is displacement. In this case, the initial velocity u = 0 (since the rock is dropped), and the acceleration a = -9.81 m/s^2 (due to gravity, acting downward).

a) To find the average velocity of the rock during its fall, we can calculate the final velocity v when the rock reaches the ground, then take the average of the initial and final velocity.

Using equation 3: v^2 = u^2 + 2as, we have:

v^2 = 0 + 2*(-9.81)*(-7)
v^2 = 137.62
v = sqrt(137.62) ≈ -11.73 m/s

(Note that the velocity is negative, meaning it is downward.)

The average velocity during the fall is:

(Initial velocity + Final velocity) / 2 = (0 + (-11.73)) / 2 = -5.865 m/s

b) To find the time it takes the rock to fall, we can use equation 2: s = ut + (1/2)at^2, with s = -7m (downward displacement) and u = 0:

-7 = 0*t + (1/2)*(-9.81)*t^2

Rearranging the equation to solve for t:

t^2 = (2 * -7) / (-9.81)
t^2 ≈ 1.428
t ≈ sqrt(1.428) ≈ 1.195 seconds

So it takes the rock approximately 1.195 seconds to fall.