You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

1 answer:

4 0

List out all the variables that you do know;

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41
v=√47.41
v=6.88549 ≈ 6.9 ms⁻¹

Hope I helped :)

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a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

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The second stone hits the ground exactly one second after the first.

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$h = v_0_yt + \frac{1}{2} gt^2$

where;

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The time taken by the first stone to hit the ground is calculated as;

$t_1 = \sqrt{\frac{2h}{g} }$

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

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