A ball is thrown vertically upwards with a speed of 10ms from bottom of a tower 200m high . Another is dropped vertically downward simultaneously from top of the tower . The time interval after which the projected body will be same level as dropped body is
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Answer:
= 20 seconds
Explanation:
We know that by second equation of newton we have
S
A
=10t−
2
1
gt
2
S
B
=
2
1
gt
2
S
A
+S
B
=200=10t
⇒t=20 sec
Verified answer
Answer :
20 sec
Explanation :
Let the time interval after which the projected body will be same level as dropped body be "t sec"
height of tower, h = 200 m
speed, u = 10 m/s
acceleration due to gravity, g =10 m/s²
The distance covered by this ball in t sec,
[tex]\boxed{\bf S_1=ut-\frac{1}{2}gt^2}[/tex]
[tex]S_1=10t-\frac{1}{2}(10)t^2 \\\\ S_1=10t-5t^2[/tex]
initial velocity, u = 0
acceleration due to gravity, g = 10 m/s²
The distance covered by this ball in t sec,
[tex]\boxed{\bf S_2=ut+\frac{1}{2}gt^2}[/tex]
[tex]S_2=0(t)+\frac{1}{2}(10)t^2 \\\\ S_2=\frac{1}{2}(10)t^2 \\\\ S_2=5t^2[/tex]
The distances covered by the both balls together is equal to the height of the tower.
S₁ + S₂ = h
10t - 5t² + 5t² = 200
10t = 200
t = 200/10
t = 20 sec
⇒ The time interval after which the projected body will be same level as dropped body is 20 sec