Question

From the top of the building 160m
high, a ball is drobbed and at the same
time a stone is projected ufowards from
the ground with a velocity of 40m/s
Find the time when ball and stone
will meet
take g as 10m/s^2

​

Answer 1

Answer:

See they will meet at a common height.

Step-by-step explanation:

speed of the stone=

$u {}^{2} \div 2g$

$40 ^{2} \div 2 \times 10$

$1600 \div 20$

$80m \div sec$

speed of ball=

$\sqrt{ \frac{2h}{g} }$

$\sqrt{ \frac{2 \times 160}{10} }$

$\sqrt{ \frac{320}{10} }$

$\sqrt{32}$

$5.656854249$

taking approx as 6

Answer 2

Answer: u1 and u2 are initial velocity of ball a and b respectively

Step-by-step explanation:

Height H is equal to S is equal to 2 is equal to 8 distance travelled by ball b s is equal to ut+ 1 upon 2 at h is

equal to who is equal to dash is equal to 40 minus C square is equal to 14 14 14 t minus 10 -10 -10 metre metre square displacement of body is equal to UT plus one upon 2 160 - height is equal to zero + 10 metre square square from first and second time is equal to 160 divided by 40 the answer will be 8ms2