Question

1 projectile motion
1.1 taking a shot (10 pts)
(problem 4.3.33)
a bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s.
(a) how much time elapses before the bullet hits the ground? (5 pts)
(b) how far does the bullet travel horizontally? (5 pts)
solutions

Explanation:

Step1: Analyze vertical motion

The bullet has vertical motion with initial vertical velocity \(v_{0y} = 0\ m/s\), acceleration \(a = g= 9.8\ m/s^{2}\) and vertical displacement \(y - y_{0}=- 1.5\ m\) (taking downwards as negative). Use the equation \(y - y_{0}=v_{0y}t+\frac{1}{2}at^{2}\). Substituting the values: \(-1.5 = 0\times t+\frac{1}{2}\times(- 9.8)t^{2}\).

Step2: Solve for time \(t\)

From \(-1.5=-4.9t^{2}\), we get \(t^{2}=\frac{1.5}{4.9}\), then \(t=\sqrt{\frac{1.5}{4.9}}\approx0.55\ s\).

Step3: Analyze horizontal motion

For horizontal motion, there is no acceleration (\(a_x = 0\)), initial horizontal velocity \(v_{0x}=200\ m/s\). Use the equation \(x - x_{0}=v_{0x}t\).

Step4: Calculate horizontal distance

Substitute \(v_{0x} = 200\ m/s\) and \(t = 0.55\ s\) into \(x - x_{0}=v_{0x}t\). So \(x - x_{0}=200\times0.55 = 110\ m\).

Answer:

(a) The time elapsed before the bullet hits the ground is approximately \(0.55\ s\).
(b) The bullet travels approximately \(110\ m\) horizontally.