Question

question 13
a projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s. how long does it take the projectile to reach the highest point in its trajectory?
6.2 s
4.0 s
2.9 s
9.8 s
1.5 s

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 35.0$ m/s and the launch angle $\theta=55.0^{\circ}$. The vertical - component of the initial velocity is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=35\sin(55^{\circ})\approx35\times0.819 = 28.665$ m/s.

Step2: Use kinematic equation for vertical motion

At the highest - point of the trajectory, the vertical velocity $v_y = 0$. The kinematic equation $v_y=v_{0y}-gt$ is used, where $g = 9.8$ m/s². We want to find the time $t$. Rearranging the equation for $t$ gives $t=\frac{v_{0y}-v_y}{g}$. Substituting $v_y = 0$ and $v_{0y}=28.665$ m/s and $g = 9.8$ m/s², we get $t=\frac{28.665 - 0}{9.8}\approx2.9$ s.

Answer:

$2.9$ s