Question

question 14 8 pts a projectile is launched with an initial speed of 56.6 m/s at an angle of 39.7 above the horizontal. the projectile lands on a hillside 3.6 s later. neglect air friction. (assume that the +x - axis is to the right and the +y - axis is up along the page.) what is the magnitude of the projectiles velocity at the highest point of its trajectory? express your answer in m/s. question 15 8 pts a projectile is launched with an initial speed of 61.6 m/s at an angle of 39.6° above the horizontal. the projectile lands on a hillside 5 s later. neglect air friction. (assume that the +x - axis is to the right and the +y - axis is up along the page.) what is the angle of the projectiles velocity at the highest point of its trajectory, relative to the horizontal (the +x - axis)?

Explanation:

Question 14

Step1: Resolve initial velocity into components

The initial velocity $v_0 = 56.6$ m/s and the launch - angle $\theta=39.7^{\circ}$. The horizontal component of the initial velocity is given by $v_{0x}=v_0\cos\theta$. At the highest point of the projectile's trajectory, the vertical component of the velocity $v_y = 0$. The velocity at the highest point is equal to the horizontal component of the initial velocity.
$v_{0x}=v_0\cos\theta=56.6\times\cos(39.7^{\circ})$

Step2: Calculate the value

Using a calculator, $\cos(39.7^{\circ})\approx0.77$, so $v_{0x}=56.6\times0.77 = 43.6$ m/s

At the highest point of the projectile's trajectory, the vertical component of the velocity $v_y = 0$. The velocity vector at the highest - point consists only of the horizontal component of the velocity. The angle of the velocity vector $\theta_v$ with respect to the horizontal is given by $\tan\theta_v=\frac{v_y}{v_x}$. Since $v_y = 0$ and $v_x=v_0\cos\theta$ (where $v_0 = 61.6$ m/s and $\theta = 39.6^{\circ}$), but we don't need to calculate $v_x$ for finding the angle.

Answer:

$43.6$

Question 15