Answer to Question #227705 in Mechanics | Relativity for nad

Question

A projectile is launched with an initial speed of 46.0 m/s at an angle of 35.0° above the horizontal. The projectile lands on a hillside 3.95 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(a) What is the projectile's velocity at the highest point of its trajectory?

Answer 1

$v_0 = 46\frac{m}{s}$

$\alpha = 35\degree$

$t_1 = 3.95s$

$g= 9.8\frac{m}{s^2}$

$\text {Projectile rise time:}$

$t_p = \frac{v_0\sin\alpha}{g}= \frac{46*\sin35\degree}{9.8}=2,69s$

$t_p<t_1\text{ that is, the projectile reaches the top point of the trajectory}$

$\text{{ without colliding with the mountainside}}$

$\text{for }t_p:$

$v_y = 0;v_x= v_0\cos\alpha;$

$v = 46*\cos35\degree=37.68\frac{m}{s}$

$\text {Answer: }v =37.68\frac{m}{s}$