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