TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given, Let height of ball from ground be H and time taken by ball l to reach on ground be t.
∴‌‌H=500m
∴ Time taken by ball 2 to reach on ground
t2=(t−1)
As we know that
H=ut+‌

1

2

gt2
where, u is initial speed i.e. 0ms−1 for free fall and g is acceleration due to gravity i.e. 10ms−2.
∴‌500=0t+‌

1

2

gt2‌‌‌ [For the first ball] ‌
⇒‌500=‌

1

2

×10t2
⇒‌100=t2
⇒‌‌t=10s
∴H2 (distance covered by ball 2)
H2=u(t−1)+‌

1

2

g(t−1)2
⇒‌‌H2=‌

1

2

g(10−1)2
[∵u=0]
=‌

1

2

×10×92
=5×81=405m
Hence, distance between the two balls, when first ball is in contact with ground is equal to ∆H, then
∆H‌‌=H1−H2
‌‌=500−405=95m