bomber on a military mission is flying horizontally at a weight of 3000m above the ground at 60km mn–¹,it drops a bomb on a target on the ground.determine the acute angle between the vertical and the line joining the bomber
asked by Lucy on August 24, 2019
The horizontal speed of the bomb is the same as the bomber speed.
In other words the bomb explodes exactly under the bomber.
( If you drop a bomb at low altitude, bank and turn immediately.)
To determine the acute angle between the vertical and the line joining the bomber to the target, we can use trigonometry.
Let's break down the given information:
- The bomber's height above the ground is 3000m.
- The bomber is flying horizontally, which means its vertical velocity component is 0.
- The bomber's speed is given as 60 km/min.
To find the angle, we can use the tangent function. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the opposite side is the height of the bomber and the adjacent side is the horizontal distance from the bomber to the target.
Step 1: Convert the speed of the bomber from km/min to m/s.
Since 1 km = 1000 m and 1 min = 60 s, we can multiply the speed by the conversion factor (1000m/60s) to get the speed in m/s.
60 km/min x (1000m/60s) = 1000 m/min = 1000/60 m/s = 16.67 m/s (approximately)
Step 2: Calculate the time it takes for the bomb to reach the ground.
The time can be calculated using the formula: time = distance/speed.
Since the height of the bomber above the ground is 3000m and the vertical velocity component is 0, the distance traveled by the bomb is also 3000m.
Therefore, the time it takes for the bomb to reach the ground is 3000m / 16.67 m/s = 180.02 s (approximately).
Step 3: Calculate the horizontal distance traveled by the bomber during the time the bomb is in the air.
The horizontal distance can be calculated using the formula: distance = speed x time.
The speed of the bomber is 16.67 m/s (from step 1) and the time is 180.02 s (from step 2).
Therefore, the horizontal distance traveled by the bomber is 16.67 m/s x 180.02 s = 3000.54 m (approximately).
Step 4: Calculate the tangent of the angle.
The tangent of the angle is equal to the ratio of the height of the bomber to the horizontal distance traveled by the bomber.
The height of the bomber is 3000m (given) and the horizontal distance is 3000.54m (from step 3).
Therefore, the tangent of the angle is 3000m / 3000.54m = 0.9998.
Step 5: Find the acute angle.
To find the acute angle, we need to take the inverse tangent (arctan) of the tangent value calculated in step 4.
The acute angle is equal to arctan(0.9998) = 44.99 degrees (approximately).
Therefore, the acute angle between the vertical and the line joining the bomber to the target is approximately 44.99 degrees.