The problem involves a ball in free fall, starting from rest. We are given the final velocity reached and the acceleration due to gravity. We need to determine the height from which the ball fell, which corresponds to the height of the building.
We can use a standard kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The appropriate equation is:
\(v^2 = u^2 + 2as\)
Substitute the known values into the kinematic equation:
\((40\text{ m/s})^2 = (0\text{ m/s})^2 + 2 \times (10\text{ m/s}^2) \times h\)
Simplify the equation:
\(1600\text{ m}^2/\text{s}^2 = 0 + (20\text{ m/s}^2) \times h\)
\(1600\text{ m}^2/\text{s}^2 = (20\text{ m/s}^2) \times h\)
Now, solve for the height (\(h\)):
\(h = \frac{1600\text{ m}^2/\text{s}^2}{20\text{ m/s}^2}\)
\(h = 80\text{ m}\)
Therefore, the height of the building is 80 meters.
A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X 1, in the first 10 s and distance X 2in the remaining 10 s, then which of the following is true?
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
If the distance travelled by a body in the $n^{th}$ second is given by $(7 + 5n)$ m, then find the initial velocity and acceleration of the body respectively.
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2