Question

A ball is thrown up caught by thrower after 4sec . How high did it go and with what velocity was it thrown?

Answer 1

Answer:

total time taken = 2u/g

as g = 10 m/s^2

and total time taken is 4 sec,

4=2u/10

2u = 40

u= 20 m/s

maximum height = u²/2g

                            =400/20

                            = 20 m

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Answer 2

Answer:

a = -g

v= u+ at

0= u -g(2)

u= 2g = 2(10) = 20m/sec

Initial velocity of ball is 20m/sec.

and because v2-u2 = 2as

0 - (20)2 = -2(10)h

h = 20m

HERE IS YOUR ANSWER MATE!

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