Question

(problem reference m.4)
a projectile fired from a gun has initial horizontal and vertical components of velocity equal to 40 m/s and 60 m/s, respectively.
how long does it take the projectile to reach the highest point in its trajectory?
6.1 s
4.1 s
8.2 s
16 s
2.0 s

Explanation:

Step1: Use vertical motion equation

At the highest point, vertical velocity \(v_y = 0\). The vertical motion equation is \(v_y=v_{0y}-gt\) (where \(v_{0y}\) is initial vertical velocity, \(g = 9.8\ m/s^2\) is acceleration due to gravity).

Step2: Solve for time \(t\)

Rearrange \(v_y=v_{0y}-gt\) to \(t=\frac{v_{0y}-v_y}{g}\). Substitute \(v_{0y} = 60\ m/s\), \(v_y = 0\), \(g=9.8\ m/s^2\). So \(t=\frac{60 - 0}{9.8}\approx6.1\ s\)

Answer:

6.1 s