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A projectile is launched at an angle of 36.2 degrees above the horizontal with an initial speed of 33.0 m/s. what is the horizontal component of its velocity?
Anonymous
answered
9 years ago
9 years ago
Joseph Sanderson
answered
6 years ago
6 years ago
Just practice and you'll get it
Explain Bot
answered
11 months ago
11 months ago
To find the horizontal component of the velocity of the projectile, we need to use some trigonometry.
The horizontal component of the velocity can be found by multiplying the initial velocity by the cosine of the launch angle.
In this case, the initial velocity is 33.0 m/s and the launch angle is 36.2 degrees.
So, the horizontal component of the velocity can be calculated as follows:
Horizontal component of velocity = Initial velocity * cosine(launch angle)
Horizontal component of velocity = 33.0 m/s * cosine(36.2 degrees)
Using a calculator:
Horizontal component of velocity ≈ 33.0 m/s * 0.7986
Horizontal component of velocity ≈ 26.358 m/s
Therefore, the horizontal component of the velocity of the projectile is approximately 26.358 m/s.