Question

part 11 of 11 - analyze
meixu 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.)
10.5 √ 10.5 s
going further: maximum flight time and angle
the friends found that the maximum time of flight occurs when the initial launch angle is 90°. joey asks meixu why this is so. which of her responses is correct?
\when the launch angle is 90°, there is no component of the velocity moving against the acceleration due to gravity, therefore, the time of flight is a maximum.\
\when the launch angle is 90°, sin(θ) takes its minimum value, so t_total is maximum.\
\for this angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction.\
\for this angle, the vertical component of velocity is zero, so the initial velocity is completely in the horizontal direction.\

Explanation:

Step1: Analyze vertical - motion equation

The time - of - flight formula for a projectile is derived from the vertical - motion equation $y = y_0+v_{0y}t-\frac{1}{2}gt^{2}$. When the projectile returns to the same height ($y = y_0$), the equation becomes $0 = v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the launch time), and the other is $t=\frac{2v_{0y}}{g}$, where $v_{0y}=v_0\sin\theta$ ($v_0$ is the initial speed and $\theta$ is the launch angle).

Step2: Consider the case of $\theta = 90^{\circ}$

When $\theta = 90^{\circ}$, the initial velocity $v_0$ has no horizontal component ($v_{0x}=v_0\cos\theta=v_0\cos90^{\circ}=0$) and the entire initial velocity is in the vertical direction ($v_{0y}=v_0\sin\theta = v_0\sin90^{\circ}=v_0$).

Answer:

"For this angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction."