Question #107629
A ball bearingX is dropped vertically downwards from the edge of a table and it takes 0.5 seconds to hit the floor.another one y leaves the edge of the table horizontally with a velocity of 5mpr second find
A.time taken for y to reach the floor
1
Expert's answer
2020-04-02T11:01:06-0400

As per the given question,

time taken to reach the ground by X one (t1)=0.5sec(t_1)=0.5 sec

we know that h=ut+gt22h=ut+\dfrac{gt^2}{2}

u=0

h=0+9.8×0.5×0.52h=0+\dfrac{9.8\times 0.5 \times 0.5}{2}

h=1.225mh=1.225m

now for the Y,

object is projected in horizontal direction, but there is no vertical velocity, so intital velocity in the vertical direction =0

so h=ut+gt22h=ut+\dfrac{gt^2}{2}

1.225=0+9.8×t22\Rightarrow 1.225=0+\dfrac{9.8\times t^2}{2}

t=0.25=0.5sect=\sqrt{0.25}=0.5 sec


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Comments

Faith
07.04.20, 00:35

Thank you so much.. My question was answered correctly and it solo nicely done..

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