Physics, 10.07.2019 04:00 ayoismeisalex

The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. what is the initial projection angle of the projectile?

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The initial projection angle of the projectile is arc tangent of the ratio of vertical velocity component to the horizontal. Angle is 70.4 degrees.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

It is given that the maximum height reached by the projectile is 0.58 times its speed when it is at half of its maximum height.

Suppose the v_1 is the speed of the projectile at the maximum height and v_2 is the speed of the projectile at the half of its maximum height

As maximum height reached by the projectile is 0.58 times its speed when it is at half of its maximum height. Therefore,

v_1=0.58v_2

Now at the maximum speed the vertical component of velocity is zero. Thus,

$\sqrt{v_x^2+0}=0.58\sqrt{v_x^2+v_y^2}\\v_x=0.58\sqrt{v_x^2+v_y^2}\\$

Squaring both sides, we get,

$v_x^2=0.3364({v_x^2+v_y^2})\\v_x^2=0.5076v_y^2$

From the third equation of motion at the distance of half of the maximum height,

$v_y^2-v_{iy}^2=2g\dfrac{h}{2}\\$
Let the above equation as equation 1.

The height of the projectile can be given as,

$h=\dfrac{v_{iy}^2}{2g}$

Put this value in the equation one, we get,

$v_y^2-v_{iy}^2=2g\dfrac{v_{iy}^2}{2\times2g}\\2v_y^2=v_{ig}$

Comparing both the values of v_y , we get,

$v_x=0.356v_{iy}$

For the angle of projection, we know that,

$\theta=\tan^{-1}\dfrac{v_{iy}}{v_x}\\\theta=\tan^{-1}\dfrac{v_{iy}}{0.356v_{iy}}\\\theta=70.4^o$

Thus, the initial projection angle of the projectile is 70.4 degrees.

Learn more about the projectile motion here;

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Answer from: NetherisIsTheQueen

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

v_1 = v_x

now when he is at half of the maximum height the speed will be in x and y direction both

$v_2 = \sqrt{v_y^2 + v_x^2}$

here it is given that

v_1 = 0.58 v_2

$v_x = 0.58\sqrt{v_x^2 + v_y^2}$

2.97 v_x^2 = v_x^2 + v_y^2

1.97 v_x^2 = v_y^2

also we know that

$v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}$

here we know that maximum height is given as

$H = \frac{v_{iy}^2}{2g}$

$v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}$

$v_y^2 = \frac{v_{iy}^2}{2}$

now from above

$1.97 v_x^2 = \frac{v_{iy}^2}{2}$

$1.98 v_x = v_{iy}$

also we know that angle of projection is

$tan\theta = \frac{v_{iy}}{v_x}$

$tan\theta = \frac{1.98v_x}{v_x}$

so angle is

$\theta = tan^{-1} 1.98$

$\theta = 63.3 degree$

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