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A boy throws an arrow at an original velocity of 2m / s, aiming to create an angle 0, referring to the balloon at a distance of 3m from the point of departure. Calculate the angle 0 and the height of the arrow.

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Answer:

s=0.204m

Step-by-step explanation:

Assuming that the arrow is thrown horizontally and there is no air resistance, we can use the following formula to calculate the launch angle 0:

tan(0) = opposite/adjacent = height/distance

where opposite is the height that the arrow reaches and adjacent is the distance to the balloon.

Rearranging the formula, we get:

0 = arctan(height/distance)

0 = arctan(height/3)

Taking the tangent of both sides, we get:

tan(0) = tan(arctan(height/3))

tan(0) = height/3

Now, we need to find the height of the arrow. Using the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (0m/s, at maximum height), u is the initial velocity (2m/s), a is acceleration (-9.8m/s^2, due to gravity) and s is the distance travelled vertically until the arrow reaches maximum height.

At maximum height, the final velocity is 0m/s. Therefore, we have:

0 = (2m/s)^2 + 2(-9.8m/s^2)s

Solving for s, we get:

s = 0.204m

Therefore, the height of the arrow is approximately 0.204m.

User Oguz Ozgul
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