Question

A ball thrown vertically upward returns back to the
thrower in 4 s. Total distance covered by the ball is
[Take, g = 10 m/s21​

Answer 1

Answer:

we know that time of descending = time if ascending

total time taken=4 then time for ascending=2sec

at highest point v=0

g=-10m/s^2(for ascending)

By using first eq of motion

v=u+gt

-u=-gt

u=10×2

u=20m/s

Max height attained

by using third eq of motion

v^2-u^2=2as

u^2=2as

20*20=2×10×s

400=20*s

400/20=s

s=20m

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