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Please help me with sum 15 16 and 17 of the attached imagePlsss

Please help me with sum 15 16 and 17 of the attached imagePlsss

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Grade:11

3 Answers

Eshan
askIITians Faculty 2095 Points
4 years ago
Dear student,

Please post one question per thread only,

Answer to question number 15-

Let the height of point from ground where they meet behand time after which they meet bet.

Hence for first ball,100-h=\dfrac{1}{2}gt^2

For the second ball,h=25t-\dfrac{1}{2}gt^2

Adding the two gives, t=4s, h=20m
Venkat
273 Points
4 years ago
Take height from the top as x
 
When dropped from top , u = 0, g = 9.8 , h = x, time taken is t
x=ut+0.5gt2 = 0 + 0.5(9.8)t2= 4.9t2   (1)
 
When the  ball projected upward, h = 100-x , u = 25 m/s , time is same t
100-x = 25 t-0.5*9.8*t2
x = 100-25t+4.9t2 (2)
 
from (1) and (2)
4.9t2 = 100-25t+4.9t2
t = 100/25 = 4 seconds
 
From (1)
x= 4.9(4)2    = 78.4 meters
Aman
16 Points
3 years ago
For 14
Let height reached by particle on the top of building be x
Initial velocity u= 14m/s
And final velocity v =0
As v= u+at
v= u- gt
t= 1.42s
As h= ut -1/2gt
h = 14×1.42- 1/2 ×10×1.42×1.42
   = 19.88- 10.08
    = 9.8m
Given total time taken= 4.6s
Time left 4.6-1.42= 3.18s
Second case::
Velocity at heighest point is zero
u=0
v =? 
t=3.18s
v= 9.8×3.18= 31.164m/s
2as= v2 - u2
s= (31.164×31.164)/2×9.8
  = 49.55m
Total height = 49.55m
Height of building = 49.55-9.8
  = 39.7m
 
 
 

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