Estimate (a) how long it took King Kong to fall straight downfrom the top of the

Question

Estimate (a) how long it took King Kong to fall straight downfrom the top of the empire state building (380m high) and (b) hisvelocity just before "landing" y=0+1/2(9.80) x t^2 2x380=9.80t^2 t^2= 380x2/9.80 t=760/9.80 t=7.75 B King Kong's velocity just before landing v=? V^2=V^2+2gh V^2=0+2(9.80)(380)          0+(19.6)(380)          0+7448 Help, I can't figure out what I'm doing wrong. Thank you inadvance for your help. Estimate (a) how long it took King Kong to fall straight downfrom the top of the empire state building (380m high) and (b) hisvelocity just before "landing" y=0+1/2(9.80) x t^2 2x380=9.80t^2 t^2= 380x2/9.80 t=760/9.80 t=7.75 B King Kong's velocity just before landing v=? V^2=V^2+2gh V^2=0+2(9.80)(380)          0+(19.6)(380)          0+7448 Help, I can't figure out what I'm doing wrong. Thank you inadvance for your help.

Explanation / Answer

Height h = 380 m

Initial speed u = 0

(a). time taken to land T = [ 2h / g ]

                                       = 8.806 s

(b). speed when it lands v = u + gT

                                          =86.3 m / s

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