Question

SA12. Illustrate the path of a projectile launched at a velocity of 150.0 m/s at an angle of 50O with the horizontal and find, illustrate, and label the following in the same illustration you have. a. The viy of the projectile

 b. The vix of the projectile

 c. Its vy 2.0 seconds after it was launched

 d. Its vf based on letter c 

e. Its maximum height (𝐦𝐚𝐱 𝐝𝐲 ↑)

 f. The time it will reach a height of 3.0 m (𝒕 ↑) 

g. Its position/location 4.0 seconds of its flight (𝐝𝐲 ↑) 

h. Its range ( R ) 

Answer 1

Explanations & Calculations

c)

• Apply v = u + at upwards,

$\qquad\qquad\small v_y=150\sin50-9.8\times2s=95.3\,m$

e)

• Apply $\small v^2=u^2+2as$ upwards for the motion of the projectile.

$\qquad\qquad \begin{aligned} \small 0^2&=\small (150\sin50)^2+2\times(-9.8)h\\ \small h&=\small 667.5\,m \end{aligned}$

f)

• Apply $\small s= ut+\frac{1}{2}at^2$ upwards,

$\qquad\qquad \begin{aligned} \small 3&=\small 150\sin50 .t+0.5(-9.8).t^2\\ \small t&=\begin{cases} \small 23.42\,s\\ \small 0.026\,s \end{cases} \end{aligned}$

• Therefore, it takes 0.026 seconds to reach 3m on its way up (for the first time).
• On its way down it takes 23.42 seconds to come back to a position 3m above the starting level.

g)

• Apply $\small s=ut+\frac{1}{2}at^2$ upwards,

$\qquad\qquad \begin{aligned} \small h_1&=\small 150\sin50\times4s+0.5(-9.8)\times4^2\\ &=\small 381.2\,m \end{aligned}$

h)

• Apply $\small s = ut$ horizontally

$\qquad\qquad \begin{aligned} \small R&=\small 150\cos50\times23.45s\\ &=\small 226.6\,m \end{aligned}$