Physics, asked by soorajjohnson28, 1 month ago

A ball is thrown up with a speed of 15 m/s. how high will it go before it begins to fall?

Answers

Answered by kargetikavita8

0

Explanation:

(v) = 0(u) = 15 m/s

a = -g = -9.8

s= (v)^2 - (u) ^2/2a

s = (0 - 225)/(2 *-9.8)

= 225/19.6

=11.47 m

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