(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°
<h3>Further explanation</h3>
Acceleration is rate of change of velocity.
<em>a = acceleration ( m/s² )</em>
<em>v = final velocity ( m/s )</em>
<em>u = initial velocity ( m/s )</em>
<em>t = time taken ( s )</em>
<em>d = distance ( m )</em>
Let us now tackle the problem!
This problem is about Projectile Motion
The complete question as follows:
<em>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 </em><em>i</em><em> + 6.1 </em><em>j </em><em>in meters per second, where </em><em>i </em><em>and </em><em>j </em><em>are the typical cartesian unit vectors. You can ignore air resistance. </em>
<em>(a) to what maximum height will the ball rise? </em>
<em>(b) what will be the total horizontal distance traveled by the ball? </em>
<em>(c) what is the velocity of the ball the instant before it hits the ground?</em>
<u>Given:</u>
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
<u>Unknown:</u>
a. maximum height = H = ?
b. horizontal distance = R = ?
c. final velocity = v = ?
<u>Solution:</u>
<h2><em>Question a:</em></h2>
<em>Let's find the initial vertical component of velocity:</em>
<em>At the maximum height of the ball , the vertical component of velocity is zero:</em>
<h2><em>Question b:</em></h2>
<em>The total horizontal distance traveled by the ball will be:</em>
<h2><em>Question c:</em></h2>
<em>Before it hits the ground , the velocity of the ball will be equal to its initial velocity.</em>
<h3>Learn more</h3>
<h3>Answer details</h3>
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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