An experimental jet plane, just after takeoff, climbs with a speed of 500 km/h at an angle of 37o with the horizontal. What are the horizontal and vertical velocities of the jet plane using trigonometric method?
To find the horizontal and vertical velocities of the jet plane, we can use trigonometry.
Given information:
Speed of the jet plane = 500 km/h
Angle with the horizontal = 37 degrees
First, we need to find the horizontal component of the velocity, which represents the velocity in the x-direction. This can be found by multiplying the speed by the cosine of the angle.
Horizontal velocity = Speed × cos(angle)
Horizontal velocity = 500 km/h × cos(37 degrees)
To calculate this, we need to convert the angle from degrees to radians, as the trigonometric functions in most programming languages use radians. The conversion from degrees to radians is given by:
radians = degrees × π/180
Using this, the horizontal velocity can be calculated as:
Horizontal velocity = 500 km/h × cos(37 × π/180)
Similarly, we can find the vertical component of the velocity, representing the velocity in the y-direction. This can be found by multiplying the speed by the sine of the angle.
Vertical velocity = Speed × sin(angle)
Vertical velocity = 500 km/h × sin(37 degrees)
Again, we need to convert the angle to radians:
Vertical velocity = 500 km/h × sin(37 × π/180)
So, using trigonometry, the horizontal and vertical velocities of the jet plane after takeoff are given by the above calculations.