Question

A BALL IS DROPPED FROM EDGE OF A ROOF .IT TAKES 0.1 SEC TO CROSS A WINDOW OF HEIGHT 2m. FIND THE BEIGHT OF ROOF ABOVE THE TOP OF WINDOW.

Answer 1

let it takes t second to reach the top of the window.
So it will take (t+0.1) s to reach the bottom of the window.
distance between them = 2m

So
(0×t + 0.5×10×(t+0.1)²) - (0×t + 0.5×10×(t)²) = 2
⇒ 5(t+0.1)² - 5t² = 2
⇒ 5(t² + 0.01 + 0.2t) - 5t² = 2
⇒ 5t² + 0.05 + t - 5t² = 5
⇒ t + 0.05 = 2
⇒ t = 5 - 0.05 = 4.95s

Distance of top of window = 0.5×10×4.95² = 5×24.5025 = 122.5m

Answer 2

Answer:

The answer is 19.2 m

Explanation:

for initial velocity of window,

a=10m/s square

s=2m

t=0.1

s=ut+1/2at square

2=u/10+1/2×1/10×1/10

2=2u+1/20

40=2u+1

u=19.6m/s

Now for whole,u=0

a=10

s=unknown

v=19.6

2as=v square-u square

20s=(19.6)square

s=38416/100×1/20

s=19208/1000

s=19.208m