Question

A ball is dropped from a jumping board of a swimming pool, which is at height of 20m.a second ball is thrown from same board after 1 sec.both balls hot water together what was initial velocity with which the second ball was thrown

Answer 1

Hey Friend, Here's Your Answer,
Height (s)=20m
Initial Velocity (u)=?
Final Velocity (v)=0m/s
gravitational force (g)=10m/s^2
Using 3rd Equation of Motion,
2gs=v^2-u^2
2*10*20=0-u^2
400=-u^2
-400=u^2
-√400=u^2
-20m/s=u
Hence, Initial Velocity=-20m/s

Answer 2

Answer:

Given:

Acceleration after 1s: $-10m/s^2$          (g)=$10ms^{-2}$

case (i)= for the first ball                     Distance of fall (s)= 20m

Initial velocity(u)= 0                             Time of fall(t)= ?

Final velocity (v)= ?

Explanation:

$s=ut+\frac12gt^{2}$

$20=\frac{1}{2}*10*t^{2}$ ( ut=0)

$20=5*t^{2}$

$\frac{20}{5}=t^{2}$

$4=t^{2}$

$2=t$

Since the second ball is thrown 1 second later. Therefore for the two balls to reach the ground at the same time, the second ball should be in motion for t = 2 - 1 = 1 second

using equation

$s=ut+\frac{1}{2}gt^{2}$

$20=u*1+\frac{1}{2}*10*1^{2}$

$20=u+5*1$

$20=u+5$

$20-5=u$

$15m/s^{-1}= u$

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