. A ball is dropped from the height 100 m. How much time it will take to fall on the ground? What is its velocity of strike on the ground?
Answers
Answer:
Step 1: Analysing the problem [Refer Figure]
Let the two ball meet at point C at time t
If the distance covered by ball 2 is x, then the distance covered by ball 1 will be 100−x, Since the total hieght is 100m.
Step 2: Equations of motion for constant acceleration
Since acceleration is constant, Therefore we can apply equations of motion for both balls (considering Positive Downwards).
For ball 1:
u
1
=0 a
1
=g=10m/s
2
s
1
=100−x
S
1
=u
1
t+
2
1
a
1
t
2
⇒ 100−x=
2
1
(10)t
2
....(1)
For ball 2:
u
1
=−25m/s a
2
=g=10m/s
2
s
2
=−x
S
2
=u
2
t+
2
1
a
2
t
2
⇒ −x=−25t+
2
1
10t
2
.
⇒ x=−25t−5t
2
....(2)
Step 3: Solving equations
From equation (1) and (2)
100−25t+5t
2
=5t
2
⇒ t=4s
Putting the value of t in equation (1)
x=100−5×(4)
2
=20m
Hence they will meet 20m above the ground after t=4s
Explanation:
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