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A boy throws an arrow at an original speed of 2m / s to create an angle 0 referring to the balloon at a distance of 3m from the departure point. Calculate the angle 0 and the height of the arrow. Let g = 10m / s2.

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Calculate the horizontal component of the velocity. The horizontal component of the velocity is given by:

v_x = v * cos(theta)

where v is the original speed of the arrow and theta is the angle of projection.In this case, v = 2 m/s and theta is unknown. Solving for theta, we get:

theta = arccos(v_x / v)

theta = arccos(2 / 2) = 45 degrees

Calculate the vertical component of the velocity. The vertical component of the velocity is given by:

v_y = v * sin(theta)

In this case, v = 2 m/s and theta = 45 degrees. Solving for v_y, we get:

v_y = 2 * sin(45 degrees) = 1.414 m/s

Calculate the time of flight. The time of flight is given by:

t = 2 * v_y / g

In this case, v_y = 1.414 m/s and g = 10 m/s^2. Solving for t, we get:

t = 2 * 1.414 / 10 = 0.283 seconds

Calculate the height of the arrow. The height of the arrow is given by:

y = v_y * t - 0.5 * g * t^2

In this case, v_y = 1.414 m/s, t = 0.283 seconds, and g = 10 m/s^2. Solving for y, we get:

y = 1.414 * 0.283 - 0.5 * 10 * 0.283^2 = 0.303 meters

Therefore, the angle of projection is 45 degrees and the height of the arrow is 0.303 meters.

User Dragan Marjanovic
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