Question

A block of ice is kept at the top of a slopping roof of a house 8.1m and 5.6 m are maximum and the minimum height of the roof from the ground if the angle of inclination is 30 degree for the roof with the horizontal , them how far from the starting point , measured horizontally does the block land ,if starts sliding down from the top

Answer 1

DUAGRAM:

Given:

$Angle (θ) = 30\degree$

$Minimum height of roof (CD)= 5.6 m$

$Maximum height of roof (AF) = 8.1 m$

Solution:

$Starting -point = A$

$Distance AB= AF − BF$

$= 8.1 − 5.6= 2.5 m$

Using conservation of energy between point Aand point C :

$m×g×(∆h)= 12×m×V^2$

$9.8×2.5 = 0.5×v^2$

$v = 7 m/s$

Thus, velocity at point $= 7 m/s$

Using equation of motion in vertical direction between points C and E :

$−h = (u_yt)− (1/2×g×t^2)$

$− 5.6 = (−7Sin30\degree)t − (9.82×t^2)$

$4.9t^2 +3.5t − 5.6 = 0$

On solving the quadratic equation, we get 

$t=0.77 seconds$

Using equation of motion in horizontal direction between points C and E 

$d = (u_x×t)DE$

$= 7×Cos30\degree×0.77$

$DE= 4.66 m$

Usinh trigonometric ratio in triangle ABC:

$Tanθ = AB/BC$

$Tan30\degree= 2.5BC$

$BC = 4.33 m$

Thus, total horizontal distance covered by block of ice = 4.33 + 4.66

$=8.99 m$

$=9m$