Batman drops his keys off a building that is 100 m tall and they fall straight down. he throws his grappling hook straight down 1 second after the keys fall to try to catch them. the initial velocity of the keys is 0 m/s and the initial velocity of the grappling hook is 10 m/s. when does the grappling hook catch the keys? (assume that the grappling hook is a simple projectile)

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aristocles aristocles · Answer 1 · 2018-09-16T03:56:54+02:00

After 1 second of drop the keys will go down by distance

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]y = \frac{1}{2}9.8*1 = 4.9 m[/tex]

speed of the keys after t = 1 s

[tex]v = v_i + at[/tex]

[tex]v = 0 + 9.8*1 = 9.8 m/s[/tex]

now the relative speed of drap with respect to key is given as

[tex]v_r = 10 - 9.8 = 0.2 m/s[/tex]

now the time taken by the drap to cover the distance between them

[tex]d = v_r * t[/tex]

[tex]4.9 = 0.2 * t[/tex]

[tex]t = 24.5 s[/tex]

in this time the key will drop to the position

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]y = \frac{1}{2}*9.8*24.5^2[/tex]

[tex]y = 3186.22 m[/tex]

so it shows larger then the height of the building

So here the drap will not able to catch the keys