Question

A ball is shot from the ground into the air. at a height of 9.1 m above the ground, the velocity of the ball is observed to be ????⃗ = 7.6????̂ + 6.1????̂ in meters per second, where ????̂ and ????̂ are the typical cartesian unit vectors. you can ignore air resistance. (a) to what maximum height will the ball rise? (b) what will be the total horizontal distance traveled by the ball? (c) what is the velocity of the ball the instant before it hits the ground?

Answer 1

(a) The maximum height is about 11.00 m

(b) The total horizontal distance is about 22.77 m

(c) The velocity of the ball before it hits the ground is about 16.53 m/s at -62.63°

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

This problem is about Projectile Motion

The complete question as follows:

A ball is shot from the ground into the air. At a height of 9.1 m above the ground, the velocity of the ball is observed to be v = 7.6 i + 6.1 j in meters per second, where i and j are the typical cartesian unit vectors. You can ignore air resistance.

(a) to what maximum height will the ball rise?

(b) what will be the total horizontal distance traveled by the ball?

(c) what is the velocity of the ball the instant before it hits the ground?

Given:

horizontal component of velocity = vx = 7.6 m/s

vertical componen of velocity = vy = 6.1 m/s

height of ball = h = 9.1 m

Unknown:

a. maximum height = H = ?

b. horizontal distance = R = ?

c. final velocity = v = ?

Solution:

Question a:

Let's find the initial vertical component of velocity:

$(v_y)^2 = (v_{oy})^2 - 2gh$

$6.1^2 = (v_{oy})^2 - 2(9.8)(9.1)$

$6.1^2 = (v_{oy})^2 - 178.36$

$(v_{oy})^2 = 6.1^2 + 178.36$

$v_{oy} = \sqrt{215.57}$

$v_{oy} \approx 14.68 \texttt{ m/s}$

$\texttt{ }$

At the maximum height of the ball , the vertical component of velocity is zero:

$(v_y)^2 = (v_{oy})^2 - 2gh$

$0^2 = 215.57 - 2(9.8)H$

$215.57 = 19.6H$

$H = 215.57 \div 19.6$

$H \approx 11.00 \texttt{ m}$

$\texttt{ }$

Question b:

The total horizontal distance traveled by the ball will be:

$R = v_xt$

$R = v_x \times \frac{2v_{oy}}{g}$

$R = 7.6 \times \frac{2 \sqrt{215.57} }{9.8}$

$R \approx 22.77 \texttt{ m}$

$\texttt{ }$

Question c:

Before it hits the ground , the velocity of the ball will be equal to its initial velocity.

$v^2 = (v_x)^2 + (v_{oy})^2$

$v^2 = 7.6^2 + 215.57$

$v^2 = 273.33$

$v = \sqrt{273.33}$

$v = 16.53 \texttt{ m/s}$

$\texttt{ }$

$\theta = \tan^{-1} \frac{-v_{oy}}{v_x}$

$\theta = \tan^{-1} \frac{-\sqrt{215.57}}{7.6}$

$\theta = \tan^{-1} -1.93$

$\theta \approx -62.63^o$

$\texttt{ }$

Learn more

• Velocity of Runner : https://brainly.com/question/3813437
• Kinetic Energy : https://brainly.com/question/692781
• Acceleration : https://brainly.com/question/2283922
• The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

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