Physics

Classical Mechanics Level 1

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

11.1m 11.4m 12.5m 10.4m

Ankit Vijay by Ankit Vijay , 22, India

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2 solutions

Divyanshu Pandey
Dec 29, 2014

apply 2as=V^2-u^2 as is the height reached by ball.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Tara Mitra
Jul 5, 2014

Use (v) 2= (u) 2+ 2aS Where v = 0 ( velocity at h max is 0) U=15 (given) a= -9.81 Implies S= 11.46

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Why u taken 15m/sec as initial velocity

محمد توصیف - 6 years, 11 months ago

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because it says in the problem a ball is thrown up with a speed of 15m/sec

Mardokay Mosazghi - 6 years, 11 months ago

On latexyfying it

v 2 v^{2} = u 2 u^{2} + 2 a . s 2a.s

Given u u =15, v v =0 and g g = 9.8 9.8

Put the values to get v v = 225 9.8 × 2 \dfrac{225}{9.8\times2} = 11.4 \boxed{11.4}

Parth Lohomi - 6 years, 6 months ago

V is velocity why did you use u?

Les Play Gamer - 5 years, 1 month ago

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