Question

A child uses a rubber band to launch a bottle cap at an angle of 32.5° above the horizontal. The cap travels a horizontal distance of 1.30 min in 1.60 s. What was the initial speed of the bottle cap, just after leaving the rubber band, in m/s?m/s

Answer 1

Given

Let the initial speed is V

Angle at which rubber band is launched = 32.5

Horizontal component of initial velocity

$\begin{gathered} v_x=v\cos \theta \\ \end{gathered}$

Time is given as t = 1.60 sec

Distance in horizontal direction = 1.30 m

Procedure

$\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{x}{v\cos \theta} \\ v=\frac{x}{t\cos \theta} \\ v=\frac{1.30m}{1.60s\cdot\cos 32.5} \\ v=0.9633\text{ m/s} \end{gathered}$

The answer would be v = 0.9633 m/s