Question

jet diving 290 km/h at an angle of 30 degrees and releases a bomb that hits the ground at a distance of 700 meters. how long is the bomb in the air? how high was the release point?

Answer 1

answer is 10 seconds and 897 meters, but how do i get there?

Answer 2

sorry i just answer it quick not detail :
va=290 km/h= 81.2 m/s
\[\alpha = 30^{0}\]
d for further distance= 700 m

\[Vy = Vo \sin \alpha - g t\]
since velocity at the maximum height = 0
so Va = 0

\[0=81.2 \sin 30^{0} - 10 (t) \]

Answer 3

0 = 81.2 x 0.5 - 10 t
t maksimum = 40.6

for t further of the distance is twice the t max = 2 x 40.6 = 81.2 seconds

high maximum : \[Y maks = (Vo ^{2} \sin ^{2}\alpha) / 2g\]
\[Y maks= ( (81.2)^{2}\sin ^{2}(30)) / 2 (10)\]
\[Y maks = 82.418 meters\]