Question

A rubber band projectile was launched by Cody at an angle of 500 from the

Answer 1

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:

$v_x = \frac{d}{t}$

where

d = 12 m is the horizontal distance travelled

t = 1.60 s is the total time of flight

Substituting,

$v_x = \frac{12}{1.60}=7.5 m/s$

We also know that the horizontal velocity is related to the initial velocity of the projectile by

$v_x = u cos \theta$

where

$\theta=50^{\circ}$ is the angle of projection

u is the initial velocity

Solving for u,

$u=\frac{v_x}{cos \theta}=\frac{7.5}{cos 50}=11.7 m/s$

2) Highest point: 4.1 m

The initial velocity along the vertical direction is:

$u_y = u sin \theta = (11.7)(sin 50)=9.0 m/s$

The motion along the vertical direction is a free fall motion, so we can use the following suvat equation

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity after the projectile has covered a vertical displacement of s

$a=g=-9.8 m/s^2$ is the acceleration of gravity

At the point of maximum height, the vertical velocity is zero:

$v_y=0$

So, if we substitute it into the equation, we can find s, the maximum height:

$s=\frac{v_y^2-u_y^2}{2a}=\frac{0-(9.0)^2}{2(-9.8)}=4.1 m$