Sebuah batu dijatuhkan dari tebing yang di bawahnya terdapat laut. Selang 3,4 s kemudian terdengar bunyi batu yang menyentuh permukaan laut. Jika cepat rambat bunyi di udara 340m / s, berapakah tinggi tebing ?
V bunyi = 340 m/s
“t=3,4 s merupakan jumlah t benda saat jatuh ditambah t bunyi ke pendengaran kita. Jadi t benda saat jatuh+t bunyi=h”
Misal : t benda = t1
t bunyi = t2
t = t1 + t2
t2 = 3,4 – t1
Ditanya : h = …?
Jawab :
h = Vbunyi x t2 = 0,5 x 9,8 x t1²
340 x (3,4 – t1) = 4,6 x t1²
1156 – 340 t1 = 4,6 t1²
0 = 4,6 t1² + 340t – 1156
Menggunakan rumus : -b ± √ (b2– 4 x a x c) / 2 x a
Hanya menggunakan plus karena apabila minus hasilnya TM (tidak memenuhi)
-340 + √ (340²– 4 x 4,6 x – 1156) / 2 x 4,6 = 0
(-340 + 370) / 9,2 = 30 / 92 = 3,26
Maka, t1 = 3,26
t2 = 3,14 – 3,26
= 0,14
JADI, h = 0,14 x 340
= 48 m
ATAU
h = 0,5 x 9,8 x (3,26)²
= 4,6 x 10,63
= 48 m
h = 48 m
Cicilia Steffi H (X-i/08)
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