1) Initial velocity: 11.7 m/s
The motion of the rubber band is a projectile motion, so it has:
- A uniform motion along the horizontal direction
- A free-fall motion along the vertical direction
Since the motion along the horizontal direction is uniform, the horizontal velocity is constant, and it can be calculated as:
[tex]v_x = \frac{d}{t}[/tex]
where
d = 12 m is the horizontal distance travelled
t = 1.60 s is the total time of flight
Substituting,
[tex]v_x = \frac{12}{1.60}=7.5 m/s[/tex]
We also know that the horizontal velocity is related to the initial velocity of the projectile by
[tex]v_x = u cos \theta[/tex]
where
[tex]\theta=50^{\circ}[/tex] is the angle of projection
u is the initial velocity
Solving for u,
[tex]u=\frac{v_x}{cos \theta}=\frac{7.5}{cos 50}=11.7 m/s[/tex]
2) Highest point: 4.1 m
The initial velocity along the vertical direction is:
[tex]u_y = u sin \theta = (11.7)(sin 50)=9.0 m/s[/tex]
The motion along the vertical direction is a free fall motion, so we can use the following suvat equation
[tex]v_y^2 - u_y^2 = 2as[/tex]
where
[tex]v_y[/tex] is the vertical velocity after the projectile has covered a vertical displacement of s
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
At the point of maximum height, the vertical velocity is zero:
[tex]v_y=0[/tex]
So, if we substitute it into the equation, we can find s, the maximum height:
[tex]s=\frac{v_y^2-u_y^2}{2a}=\frac{0-(9.0)^2}{2(-9.8)}=4.1 m[/tex]
