A particle is projected upwards with an initi | vedclass

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Question

(A) The equation of motion for a particle under gravity is given by $h = ut - \frac{1}{2}gt^2$.Here,$u = 80 \; ft/sec$,$h = 96 \; ft$,and $g = 32 \; f

Vedclass · Accepted Answer

$2.0 \; sec$ and $3.0 \; sec$

Vedclass · Answer

(A) The equation of motion for a particle under gravity is given by $h = ut - \frac{1}{2}gt^2$.Here,$u = 80 \; ft/sec$,$h = 96 \; ft$,and $g = 32 \; ft/sec^2$.Substituting these values into the equation:$96 = 80t - \frac{1}{2}(32)t^2$$96 = 80t - 16t^2$Dividing the entire equation by $16$:$6 = 5t - t^2$Rearranging into a standard quadratic form:$t^2 - 5t + 6 = 0$Factoring the quadratic equation:$(t - 2)(t - 3) = 0$Thus,$t = 2 \; sec$ or $t = 3 \; sec$.The particle reaches a height of $96 \; ft$ at $2 \; sec$ while moving upwards and at $3 \; sec$ while moving downwards.

$A$ particle is projected upwards with an initial velocity of $80 \; ft/sec$. The particle will be at a height of $96 \; ft$ from the ground after:

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