# An arrow is shot at 19 degrees above the horizontal. Its velocity is 28.3 m/s. What is the maximum height (in meters) the arrow will attain (neglect air resistance)?

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The arrow is shot at 28.3 m/s in a direction making an angle of 19 degrees with the horizontal. The initial velocity of the arrow can be divided into two components, a vertical component of 28.3*sin 19 that acts in the direction vertically upwards and 28.3*cos 19 that acts in the horizontal direction.

As the arrow moves there is an acceleration acting on it due to the gravitational field of the Earth of 9.8 m/s^2 in the direction vertically downwards. This decreases the vertical component of the velocity and it is reduced to 0 at the highest point of the arrow's trajectory.

Using the formula v^2 - u^2 = 2*a*s gives the maximum height s. Substituting the values given:

0 - (28.3*sin 19)^2 = -2*9.8*s

=> s = 4.33

**The maximum height of the arrow is 4.33 m**