Question

A granite block starts sliding on a slope (inclination of 30° with the horizontal) under the effect of gravity only, along the true direction of inclination of the slope and hits the ground in 4 seconds. Considering zero friction and zero cohesion during sliding, the vertical height of the point (with respect to the ground) from where the block was dislodged is _____m. (g = 10 m/s²) (In integer)

Answer 1

Correct Answer: 20

Given a slope with an inclination of 30° and a granite block sliding down under gravity, we need to find the vertical height from which the block was dislodged. Considering gravity \( g = 10 \, \text{m/s}^2 \) and neglecting friction and cohesion, we use the following approach to solve the problem.

Since the inclination is 30°, the acceleration of the block along the slope is given by:

\( a = g \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s}^2 \)

The block hits the ground in 4 seconds. We use the kinematic equation for the distance \( s \) traveled along the slope:

\( s = ut + \frac{1}{2}at^2 \)

Assuming initial velocity \( u = 0 \), we have:

\( s = \frac{1}{2} \times 5 \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \, \text{m} \)

Next, we calculate the vertical height \( h \) from which the block was dislodged:

\( h = s \sin(30^\circ) = 40 \times \frac{1}{2} = 20 \, \text{m} \)

The computed vertical height is 20 m.