A particle is projected vertically upwards from a point O with velocity u sure that if u^2 > 2gh, the particle is at a height h above O at times which are separated by [2(u^2-2gh)1/2]/g
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Expert's answer
2018-06-08T07:44:08-0400
The particle is at a height h above O can be seen as projectile from that point with: u^'. 〖u^'〗^2=u^2-2gh. u^'=√(u^2-2gh). The time of flight of projectile is t^'=(2u^')/g=(2√(u^2-2gh))/g
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