Question

13. A ball is thrown up with a 1 speed of 15 m/s , how high will it go before it begins
to fall? (g=9.8m/s2).​

Answer 1

Step-by-step explanation:

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? Use g=9.8m/s².

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8 Answers



Kshitij Kumar Ummat

Answered September 1, 2017

Final speed at the top most point (v) = 0

Initial speed (u) = 15 m/s

a = -g = -9.8

Hence, as per the third law of motion,/

v square = u square + 2.a.s

or distance s = (v square - u square)/2.a

or s = (0 - 225)/(2 x -9.8)

or s = 225/19.6 = 11.48 mtrs

please mark my answer as brainliest

Answer 2

Given :

A ball is thrown up with a a speed of 15 m/s ,

To find :

how high will it go before it begins to fall

Solution :

➸ u = 15 m/s

➸ v = 0

➸ g = - 9.8 m/s²

➸ v² - u² = 2gh .............. (Formula)

➸ h = v² - u²/2g

➸ h = 0² - 15²/2 x - 9.8

➸ h = 0 - 225/- 19.6

➸ h = 11.48

Some other Formulas :

• Speed = d/t

• F = m. a

• a (bar) = v - v_0/t = Δv/Δt

• w = F x d

• P = W/Δt

• V = 1R

• PE = mgh

• p = mv