The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was 45° and neglecting air resistance, answer the following. (a) Determine the initial speed of the projectile. 28.194 Correct: Your answer is correct. m/s (b) Determine the total time the projectile was in flight. 4.068 Correct: Your answer is correct. s (c) Qualitatively, how would the answers change if the launch angle were greater than 45°

Respuesta :

(a) 28.19 m/s

The range of a projectile is given by the equation:

[tex]d=\frac{u^2 sin 2\theta}{g}[/tex]

where

u is the initial speed

[tex]\theta[/tex] is the angle of projection

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

For the projectile in this problem, we know that:

[tex]d = 81.1 m[/tex] (range)

[tex]\theta=45^{\circ}[/tex]

So we can re-arrange the equation to find the initial speed:

[tex]u=\sqrt{\frac{dg}{sin 2\theta}}=\sqrt{\frac{(81.1)(9.8)}{sin(2\cdot 45)}}=28.19 m/s[/tex]

(b) 4.068 s

The initial vertical velocity of the projectile is

[tex]u_y = u sin \theta[/tex]

and we know that at the highest point of the trajectory, the vertical velocity becomes zero, so we have

[tex]v_y = u_y - gt = 0[/tex]

which means that the time needed to reach the maximum height is

[tex]t=\frac{u_y}{g}[/tex]

The time of flight is actually twice this time, so

[tex]t=\frac{2u sin \theta}{g}[/tex]

And by substituting

u = 28.19 m/s

[tex]\theta=45^{\circ}[/tex]

[tex]g=9.8 m/s^2[/tex]

we find:

[tex]t=\frac{2(28.19) sin 45^{\circ}}{9.8}=4.068 s[/tex]

(c) If the angle increases, the range decreases and the time of flight increases

We can answer by looking at the two formulas:

[tex]d=\frac{u^2 sin 2\theta}{g}[/tex]

[tex]t=\frac{2u sin \theta}{g}[/tex]

Let's start by analyzing the range, d. We see that:

- If the angle increases, the factor [tex]sin 2 \theta[/tex] decreases (because [tex]\theta=45^{\circ}[/tex] is the angle of maximum for the sine)

- If the angle decreases, the factor [tex]sin 2 \theta[/tex] also decreases

Therefore [tex]\theta=45^{\circ}[/tex] is the angle that gives the maximum range, so any change in this will decrease the range.

Now let's analyze the time of flight, t:

- If the angle increases, the factor [tex]sin \theta[/tex] increases, so the time of flight increases

- If the angle decreases, the factor [tex]sin \theta[/tex] decreases, so the time of flight will decrease

ACCESS MORE