Question

100 Points!

A projectile is fired in the earth's gravitational field with a horizontal velocity of v=9.00 m/s. How far does it go in the horizontal direction in 0.550s? Show your work.

B) How far does the projectile go in the vertical direction in 0.550s. Show your work

Answer 1

A) 4.95 m

B) 1.48225 m

Step-by-step explanation:

Constant Acceleration Equations (SUVAT)

`\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+(1)/(2)at^2\\\\ s&=\left((u+v)/(2)\right)t\\\\v^2&=u^2+2as\\\\s&=vt-(1)/(2)at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^(-1)$\\\\$v$ = final velocity in ms$^(-1)$\\\\$a$ = acceleration in ms$^(-2)$\\\\$t$ = time in s (seconds)\end{minipage}}`

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Consider the horizontal and vertical motion of the projectile separately.

Part A

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

`u=9.00\quad v=9.00 \quad a=0\quad t=0.550`

`\begin{aligned}\textsf{Using} \quad s & = \left((u+v)/(2)\right)t:\\\\s&= \left((9+9)/(2)\right)(0.550)\\s&= (9)(0.550)\\ \implies s&= 4.95\:\sf m\\\end{aligned}`

Part B

As the projectile is fired horizontally, the vertical component of its initial velocity is zero.

Acceleration due to gravity = 9.8 ms⁻²

Resolving vertically, taking ↓ as positive:

`u=0\quad a=9.8\quad t=0.550`

`\begin{aligned}\textsf{Using} \quad s & = ut+(1)/(2)at^2:\\\\s&= (0)(0.550)+(1)/(2)(9.8)(0.550)^2\\s&= 0+(4.9)(0.3025)\\\implies s&= 1.48225\:\sf m\\\end{aligned}`