Calculate the elevation angle of a projectile

Help with physics problem and solution

The launching speed v₀ of a projectile is two times the speed of this projectile at its maximum height. Calculate the elevation angle at launching.

Given:

Ratio of launching speed to speed at maximum height.

Solution.

At maximum height the projectile has only a horizontal component of its speed. The vertical component equals zero. Check this statement in “Projectile motion” paragraph.

The horizontal component of the speed v_x is constant during the projectile motion, so at the maximum height it is the same as at launching which can be written as

v_x(max) = v_0x       (1)

Elevation at launching can be calculated from the formula

cosα = v_ox/v₀         (2)

Taking into account formula (1), the formula (2) can be rearranged to

cosα = v_ox(max)/v₀               (3)

and with the help of inverse trigonometric functions we find

α = cos^-1(v_ox(max)/v₀)    (4)

As       

v₀/ v_ox(max)= 2.0  

α = cos^-1(1/2.0)

α = 60^o    

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