Can anyone please help me to solve this problem? I am stuck! A projectile is fired from ground level with an initial speed and an angle of elevation of 30 degrees. Acceleration due to gravity is $9.8 \, m/s^2$.a) The range of the particle is $\boxed{\phantom{0}} \, \text{meters}.$b) The maximum height of the projectile is $\boxed{\phantom{0}} \, \text{meters}.$c) The speed with which the projectile hits the ground is $\boxed{\phantom{0}} \, \text{m/sec}.$ A projectile is fired from ground level with an initial speed and an angle of elevation of 30 degrees. Acceleration due to gravity is 9.8 m/s².a) The range of the particle is [ ] meters.b) The maximum height of the projectile is [ ] meters.c) The speed with which the projectile hits the ground is [ ] m/sec.

Answered: A fired from ground level initial speed…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Can anyone please help me to solve this problem? I am stuck!

A projectile is fired from ground level with an initial speed and an angle of elevation of 30 degrees. Acceleration due to gravity is $9.8 \, m/s^2$.

a) The range of the particle is $\boxed{\phantom{0}} \, \text{meters}.$

b) The maximum height of the projectile is $\boxed{\phantom{0}} \, \text{meters}.$

c) The speed with which the projectile hits the ground is $\boxed{\phantom{0}} \, \text{m/sec}.$

A projectile is fired from ground level with an initial speed and an angle of elevation of 30 degrees. Acceleration due to gravity is 9.8 m/s².

a) The range of the particle is [ ] meters.

b) The maximum height of the projectile is [ ] meters.

c) The speed with which the projectile hits the ground is [ ] m/sec.

Expert Solution

Step 1: Finding the time of flight

Given

Angle of projection $θ = 30 °$

Let the initial velocity $u$

Resolving the velocity $u$ into its $x$ and $y$ components as,

$u_{x} = u \cos (θ) u_{y} = u \sin (θ)$

We know that

$v = u - g t s = u t - \frac{1}{2} g t^{2} v^{2} - u^{2} = 2 g s$

The projectile's total duration in the air, from the moment it is launched until it touches down, is known as its time of flight. We are aware that the highest point's velocity is zero. The highest point in our scenario is A, where velocity in the $y$ direction is zero. The equation of motion in this instance can be written as,

$v_{y} = u_{y} - g t \Rightarrow 0 = u \sin (θ) - g t \Rightarrow t = \frac{u \sin (θ)}{g}$

The projectile will now return to earth in the same amount of time as before. As a result, the time of flight $(T)$ can be calculated as being equal to the projectile's travel time multiplied by two to reach its maximum point. the duration of the flight will be

$T = \frac{2 u \sin (θ)}{g}$