Question

part 11 of 11 - analyze meixiu and joey used what they have learned to solve the following problem. a projectile is launched with a launch angle of 55° with respect to the horizontal direction and with initial speed 63 m/s. how long does it remain in flight? (enter your answer in s.) enter a number submit skip (you cannot come back) resources read it

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 63$ m/s and the launch angle $\theta=55^{\circ}$. The vertical - component of the initial velocity is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=63\sin(55^{\circ})\approx63\times0.819 = 51.6$ m/s.

Step2: Use vertical - motion equation

The equation for vertical displacement $y - y_0=v_{0y}t-\frac{1}{2}gt^2$. When the projectile returns to the same height ($y - y_0 = 0$), we have $0 = v_{0y}t-\frac{1}{2}gt^2$. Factoring out $t$, we get $t(v_{0y}-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the launch time). The other non - zero solution is $t=\frac{2v_{0y}}{g}$.

Step3: Calculate the time of flight

We know that $g = 9.8$ m/s² and $v_{0y}\approx51.6$ m/s. Then $t=\frac{2\times51.6}{9.8}\approx10.5$ s.

Answer:

$10.5$