A World War II-era dive bomber is being used to drop food and supplies to outposts on remote mountaintops. During one such drop, the bomber descends with velocity v directed at an angle of 36.1° below the horizontal. The cargo is released at an altitude of 2430 m and reaches its intended drop zone with a displacement Δr of 3030 m. What is the speed of the bomber when it releases its cargo?
displacement = sqrt(x^2 + y^2)
3030^2 = x^2 + 2430^2
x = 1810 m
In Horizontal,
initial velocity, v0x = v cos36.1
a = 0
x = v0x * t
1810 = (v cos36.1) t
v t = 2240 ........(i)
In vertical,
initial velocity, v0y = - v sin36.1
ay = - 9.8 m/s^2
Applying, y = v0y t + ay t^2 / 2
- 2430 = - (v sin36.1)t - 9.8 t^2 /2
4.9 t^2 + ( v t ) sin36.1 = 2430
putting v t from (i)
4.9 t^2 + 2240 sin36.1 = 2430
4.9 t^2 =1110
t = 15.05 sec
v = 2240/15.05 = 149 m/s .......Ans
A World War II-era dive bomber is being used to drop food and supplies to outposts...
–/1 pointsSerPSE7 4.P.018. Ask Your Teacher My Notes Question PartSubmissions Used A dive bomber has a velocity of 280 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.15 km. Find the angle θ. °
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