COLLEGE PHYSICS
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ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 3, Problem 29QAP
To determine
The initial speed of the ball as it left Reggie's bat.
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You buy a plastic dart gun, and being a clever physics
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up, and it takes 5.6 s for the dart to land back at the
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What is the maximum horizontal range of your gun?
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Chapter 3 Solutions
COLLEGE PHYSICS
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- Mayan kings and many school sports teams are named for the puma, cougar, or mountain lionFelis concolorthe best jumper among animals. It can jump to a height of 12.0 ft when leaving the ground at an angle of 45.0. With what speed, in SI units, does it leave the ground to make this leap?arrow_forward• From ta = -25 s until to = -15 s the plane is taxiing east along the runway at a constant speed of 4 m/s. From ts until te = -10 s the plane slows down at a constant rate and comes to rest. • From te until to =0 the plane turns around, turning nearly in place so that by to it is facing west. • At to = 0 the plane is near the east end of the runway, at rest. From to until ti = 5.0 s it speeds up with a constant acceleration going west. At t, it is going at a speed of 8.75 m/s. • After t, the plane continues speeding up, but at a steadily reducing rate (i.e. it is still speeding up, but the magnitude of its acceleration is decreasing). It lifts off at t2 = 65 s going at a speed of 85 m/s, having gone a distance of 2980 m down the runway. • From t2 until t3 = 240 s the plane is climbing at an angle of 5.0° above horizontal, speeding up at a constant rate. At t3 it is going at a speed of 230 m/s. • At t3 the plane levels off, so it is flying at a constant height. From t3 until ta = 270 s…arrow_forwardQUESTION 6 Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y = vosin(e) а) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y Then = Vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: Thus, the time to reach the maximum height is tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt? if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ayt2 substituting, the vinitial-y expression above,…arrow_forward
- 6arrow_forwardI need help with this questionarrow_forwardSample problem in which the projectile is launched with initial horizontal velocity only and then undergoes projectile motion: 1. A child travels down a water slide, leaving it with a velocity of 4.2 m/s horizontally. The child then experiences projectile motion, landing in a swimming pool 3.2 m below the slide. a) For how long is the child airborne? b) Determine the child's horizontal displacement while in the air. c) Determine the child's velocity upon entering the water.arrow_forward
- Problems • 1. A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal with magnitude 5.0 m/s. Find the rider velocity and position after 1/4seconds.arrow_forwardcould you please provide explanationsarrow_forwardBrownie, a dog, running at a constant speed of 1.36 m/s, decided to run towards the center of the road. After 1.5 s, a parked car started moving at a constant rate of 3.1 m/s? perpendicular to Brownie's path, towards the center of the road. If their shortest distance is initially 155.6 m apart • Prove by calculation if Brownie will be hit by the car after 1 minute or not. • If Gavin, Brownie's owner, is standing 1.5 m behind the dog, at what acceleration should he run to catch Brownie after 5 seconds? (Ans.: 0.66 m/s2)arrow_forward
- Chapter 3 learning guidelinesarrow_forward#3 SOLUTION SET UP We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then o = yo = 0, vo cos 30.0° = (20.0 m/s)(0.866) = 17.3 m/s, and vo sin 30.0° = (20.0 m/s)(0.500) = 10.0 m/s %D SOLVE We first ask when (i.e., at what value of t) the ball is at a height of 3.05 m above the ground; then we find the value of x at that time. When that value of x is equal to the distance d, the ball is just barely passing over the crossbar. To find the timet when y = 3.05 m, we use the equation y = Yo + Voy Substituting numerical values, we obtain 3.05 m = (10.0 m/s)t - (4.90 m/s²)t² This is a quadratic equation; to solve it, we first write it in standard form: (4.90 m/s²)t² – (10.0 m/s² )t + 3.05 m = 0. Then we use the quadratic formula. We get 2(4.9 m/s²) x (10.0 m/s + V (10.0 m/s)² – 4(4.90 m/s² )(3.05 m) | 0.373 s,…arrow_forwardA.Which of the following vectors best describes the direction of the acceleration at point 60m,-45m ? A,B,C,or D B. Which of the following vectors best describes the direction of the horizontal velocity at point 20m, -5m ? A,B,C,orDarrow_forward
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Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY