Question

100 Points!

A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically and then strikes the floor. How long is the projectile in the air? Show your work.

B. What is the original velocity of the projectile described? Show your work.

Answer 1

A) 0.303 s (3 s.f.)

B) 6.76 ms⁻¹ (3 s.f.)

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

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:

`s=0.450\quad u=0\quad a=9.8`

`\begin{aligned}\textsf{Using} \quad s & = ut+(1)/(2)at^2:\\\\0.450&= (0)t+(1)/(2)(9.8)t^2\\0.450&= 4.9t^2\\t^2&= (0.450)/(4.9)\\t&=\sqrt{(0.450)/(4.9)\\t &=0.3030457634...\\ \implies t&=0.303\:\sf s\;\;(3\:s.f.)\\\end{aligned}`

Part B

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

Resolving horizontally, taking → as positive:

`s=2.050 \quad a=0 \quad t=0.3030457634...`

`\begin{aligned}\textsf{Using} \quad s & = ut+(1)/(2)at^2:\\\\2.050&= u(0.3030457634...)+(1)/(2)(0)(0.3030457634...)^2\\2.050&= 0.3030457634...u\\u&= (2.050)/(0.3030457634...)\\ u &=6.764654873...\\\implies u & = 6.76\:\sf ms^(-1)\;\;(3\:s.f.)\end{aligned}`