Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
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Answer:
Explanation:
Remark
The only thing that might trip you up is what to do with the angle. The vertical component of the 15 degrees does no work against anything. So the 15 degrees limits the horizontal force.
The formula is
Work = F * d * cos(15)
The givens are
F = 2000 N
d = 30 m
Cos(15) = 0.9659
Solution
Work = 2000 * 30 * cos(15)
Work = 57,955
Rounded to two places would be 5.8 * 10^4
C
