Question

a beanbag is thrown from a window 10.0 m above the ground with an initial horizontal velocity of 3.0 m/s.

(a) how long will it take the beanbag to reach the ground? that is, what is its time of flight?

(b) how far will the beanbag travel horizontally? that is, what is its range?

Explanation:

Step1: Calculate time of flight (vertical motion)

Use the equation \( y = v_{0y}t+\frac{1}{2}gt^{2} \). Since \( v_{0y} = 0\) (initial horizontal velocity), \(y = 10.0\ m\), and \(g= 9.8\ m/s^{2}\). The equation becomes \(y=\frac{1}{2}gt^{2}\), so \(t=\sqrt{\frac{2y}{g}}\).
Substitute \(y = 10.0\ m\) and \(g = 9.8\ m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times10.0}{9.8}}\)

Step2: Calculate horizontal range (horizontal motion)

Use the equation \(x = v_{0x}t\). We know \(v_{0x}=3.0\ m/s\) and we found \(t\) from step 1.

Answer:

(a) \(t=\sqrt{\frac{2\times10.0}{9.8}}\approx1.43\ s\)
(b) \(x = 3.0\times1.43 = 4.29\ m\)