Question

an object is thrown with initial speed V metre per second with an angle of projection 30 degree . what is the height and reach reached by the particle ?​

Answer 1

Answer:

Initial speed (U) : 5 m/s

angle of projectile (a) : 30°

max. hight = (U² sin² a )/2g

= (5² sin² 30°)/2 X 9.8

=[25 (1/2)²] /19.6

= (25/4)/19.6

=6.25/19.6

= 0.318 m

range = (U² sin 2a)/g

=[5² sin (2 X 30)]/9.8

=(25 sin 60)/9.8

=[25(√3/2)]/9.8

=[25(1.7/2)]/9.8

=(25 X 0.85/2)9.8

=(21.25/2)9.8

= 10.62/9.8

=1.08 m

hance, 0.318m hight and 1.08m range it will cover.

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Answer 2

Hi mate

Refer the attachment

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