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?

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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