Question

1. zoe is setting up a track for a toy car. the track has a ramp that is 32° above horizontal. if zoe wants the car to travel as a projectile for 1.0 seconds, how fast does the toy car need to be moving as it leaves the ramp? 9.2 m/s 4.0 m/s 7.4 m/s m 1.0 m/s

Explanation:

Step1: Analyze vertical motion

The vertical component of the initial velocity is \(v_{0y}=v_0\sin\theta\). Using the kinematic equation \(y = v_{0y}t-\frac{1}{2}gt^2\). When the car returns to the same - height (assuming the ramp height is negligible for the projectile - motion time calculation, \(y = 0\)). So \(0=v_0\sin\theta\times t-\frac{1}{2}gt^2\). Since \(t
eq0\) (non - zero time of flight), we can cancel out \(t\) from the equation. We get \(v_0\sin\theta=\frac{1}{2}gt\).

Step2: Solve for \(v_0\)

We know that \(\theta = 32^{\circ}\), \(t = 1.0\ s\), and \(g = 9.8\ m/s^{2}\).
From \(v_0\sin\theta=\frac{1}{2}gt\), we can solve for \(v_0\):

$$v_0=\frac{gt}{2\sin\theta}$$

Substitute the values: \(g = 9.8\ m/s^{2}\), \(t = 1.0\ s\), \(\sin32^{\circ}\approx0.53\)

$$v_0=\frac{9.8\times1.0}{2\times0.53}=\frac{9.8}{1.06}\approx9.2\ m/s$$

Answer:

9.2 m/s