Question

A block of ice starts sliding down from the top of an inclined roof of a house along a line of the
greatest slope. The inclination of the roof with the horizontal is 30°. The heights of the highest and
lowest points of the roof are 8.1 m and 5.6 m respectively. At what horizontal distance from the
lowest point will the block hit the ground? Neglect air friction. (g = 9.8 m/s']​

Answer 1

Answer:

When the ice block silides down from A to B, it falls vertically through AC=2.5m

If v is the velocity with which it reaches B, then v2=u2+2as, where u=0,a=gsin30∘ and s=AB.

∴v2=2g×ACAB×AB(sin30∘=ACAB)

=2×9.8×2.5=49

∴v=7ms−1 along AB, Resolve this velocity v into:

(i) a horizontal component

=vcos30∘=7×(3–√/2)=3.5×3–√ms−1

Vertical motion:

Let t be the time taken by the block to reach the ground.

Apply s=ut+12gt2

Given s=5.6m,u=3.5ms−1,g=9.8ms−2

∴5.6=3.5t+12×9.8t2 or 4.9t2+3.5t−5.6=0

Dividing this equation by 0.7, we get 7t2+5t−8=0

∴t=−5+0(5)2+(4×7×8)−−−−−−−−−−−−−−−√2×7

∴t=−5±15.7814

Rejecting the -ve value,

=−5+15.7814=10.7814=0.77s

Horizontal Motion:

Horizontal distance travelled EF

=Horizontal velocity xx time

=3.5×3–√×0.77=4.668m

∴ Total horizontal distance =DF=DE+EF

=4.33+4.668=8.998m

(DE=BC=ACcot30∘=2.5×3–√=4.33m)

Explanation: