Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
With an initial velocity of 23.4 m/s, an arrow is fired straight up from the ground then it will take 2.35 seconds to reach its maximum height.
What is meant by initial velocity?
Initial Velocity, denoted as u, is the velocity at time period t = 0. It is the speed at which motion first occurs. There are four equations for starting velocity: (1) The starting velocity is expressed as u = v - at if time, acceleration, and final velocity are given.
Given: The initial velocity of an arrow is 23.4 m/s
We need to find the time taken by it to reach its maximum height. Let the maximum height is h.
Using third equation of motion to find height first.
v² - u² = 2ah
At maximum height, v = 0
a = -g
u² = 2gh
h = u²/2g
substituting the values in the above equation, we get
h = (23.4)² / 2(9.8)
h = 27.93
The value of h = 27.93
Let t is the time to reach its maximum height. So, it can be calculated using second equation motion as follows:
27.93 = 23.4 t - 1/2(9.8) t²
t = 2.35 s
Therefore, it will take 2.35 seconds to reach its maximum height.
To learn more about initial velocity refer to:
https://brainly.com/question/21010554
#SPJ4
Answer:
2.39 s (2 d.p.)
Explanation:
Constant Acceleration Equations (SUVAT)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{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}}[/tex]
When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
When the arrow reaches its maximum height, the vertical component of its velocity will momentarily be zero.
The arrow is moving freely under gravity so:
[tex]a = \text{g} = 9.8 \; \sf ms^{-2}[/tex]
Resolving vertically, taking ↑ as positive:
[tex]u=23.4 \quad v=0 \quad a=-9.8[/tex]
[tex]\begin{aligned}\textsf{Using} \quad v&=u+at:\\\\0&=23.4+(-9.8)t\\0&=23.4-9.8t\\9.8t&=23.4\\t&=\dfrac{23.4}{9.8}\\ t&=2.387755102\\\implies t&=2.39 \; \sf s\;(2\:d.p.)\end{aligned}[/tex]
Therefore, the arrow will reach its highest point at 2.39 s.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
