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Grade 11Modern Physics

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

Question image for Please help me with sum 15 16 and 17 of the attach

Unknown guy

8 Years agoGrade 11

3 Answers

Eshan

8 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 be $h$ and time after which they meet be $t.$

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

8 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

7 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 -1/2gt²

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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