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: Analyze vertical motion

The vertical motion of the bean - bag is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), the vertical displacement \(y - y_{0}=- 10.0\ m\) (taking downwards as negative), and the acceleration \(a = g=-9.8\ m/s^{2}\). Use the equation \(y - y_{0}=v_{0y}t+\frac{1}{2}at^{2}\). Substituting the values, we get \(-10.0 = 0\times t+\frac{1}{2}\times(-9.8)t^{2}\).

Step2: Solve for time \(t\)

From \(-10.0=-4.9t^{2}\), we can solve for \(t\). First, rewrite the equation as \(t^{2}=\frac{10.0}{4.9}\). Then \(t=\sqrt{\frac{10.0}{4.9}}\approx1.43\ s\).

Step3: Analyze horizontal motion

In the horizontal direction, there is no acceleration (\(a_x = 0\)), so the horizontal velocity \(v_x\) is constant. The horizontal velocity \(v_x = 3.0\ m/s\), and the time of flight \(t\approx1.43\ s\). Use the equation \(x - x_{0}=v_{x}t\).

Step4: Calculate the horizontal range

Substitute \(v_x = 3.0\ m/s\) and \(t\approx1.43\ s\) into \(x - x_{0}=v_{x}t\). Then \(x - x_{0}=3.0\times1.43 = 4.29\ m\).

Answer:

(a) \(t\approx1.43\ s\)
(b) \(x - x_{0}\approx4.29\ m\)