Question

a soccer ball is kicked at an angle of 35° and it lands on even ground.
A) What angle will produce the same range?
B) Which angle of the two will produce the highest ball?
C) Which angle of the two will produce the ball that is in the air the longest?
D) What angle in this situation would produce the furthest range?

Answer 1

Hi there!

A)

The angle that will produce the same range is the compliment of 35°.

Thus, kicking the ball at 55° will result in the same range.

We can prove this by using the derived range equation:

$R = \frac{v^2sin2\theta}{g}$

An angle of 35° yields:

$R = \frac{v^2sin(2*35)}{g} = .939R$

An angle of 55° yields:

$R = \frac{v^2sin(2*55)}{g} = .939R$

Both are the same, thus indicating that 55° produces the same range.

B)

The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.

sin(55) > sin(35)

C)

The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.

We can prove this using:

vf = vi + at

0 = vy - 9.8t

vy/9.8 = t

Greater vy (vertical component of velocity) ⇒ greater time taken.

D)

The angle that would result in the furthest range is 45°.

We can prove this using calculus. Recall the above range equation:

$R = \frac{v^2sin2\theta}{g}$

We can take the derivative and use the first-derivative test to find its critical point:

$\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0$

Evaluate:

$v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}$