An observer standing near a window 5 m high observe that an object falling downwards is passing across the window in 0.5 s. Find at what height above the window the object was dropped. (g = 10 m/s2).​

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Answer:

Explanation:

An object in free fall, NOT experiencing parabolic motion, has an equation of

[tex]h(t)=\frac{1}{2} gt^2+h_0[/tex] which says:

The height of an object with respect to time in seconds is equal to the pull of gravity times time-squared plus the height from which it was dropped. Normally we use -9.8 for gravity but you said to use 10, so be it.

For us, h(t) is 5 because we are looking for the height of the window when the object is 5 m off the ground at .5 seconds;

g = 10 m/s/s, and

t = .5sec

[tex]5=\frac{1}{2}(-10)(.5)^2[/tex]+h and

5 = -5(.5)² + h and

5 = -5(.25) + h and

5 = -1.25 + h so

h = 6.25

That's how high the window is above the ground.

Height above the window object was dropped is 6.25 m

Given:

Height of window = 5 m

Time taken to drop object = 0.5 sec

Gravitational acceleration = 10 m/s²

Find:

Height above the window object was dropped

Computation:

We know that;

[tex]H(T) = \frac{1}{2}(g)(t^2) + H(S)\\\\5 = \frac{1}{2}(10)(0.5^2) +H(S)\\\\5 = (5)(0.5^2)+H(S)\\\\H(S) = 6.25[/tex]

So,

Height above the window object was dropped = 6.25 m

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