engage https://edugen.wileyplus.com/edugen/student/ma l Help I Contact Us Les 9u

Question

engage https://edugen.wileyplus.com/edugen/student/ma l Help I Contact Us Les 9ut Halliday, Fundamentals of Physics, 10e University P hysics (PHYS 07/ PHYS 108/ PHYS 207) ice Assignment Gradebook ORION Downloadable eTextbook ment MESSAGE PAY INSTRUCTOR FULLSCREEN PRINTER VERSION BACK NEXT Chapter 04, Problem 036 During a tennis match, a player serves the ball at 29.5 m/s, with the center of the ball leaving the racquet horizontally court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) w the distance between t before bu distance between the center of the ball and the top of the net? Enter a positive number if the bal ball does not clear the net, your answer should be a negative number, Use g-9.81 m/s? t now it leaves the racquet at 5.00 below the horizontal. When the ball reaches the net, what now is the (a) Number Units (b) Number Units GO TUTORIAL LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO M1 ATH HELIP Version 4.24.9 olic 12000-2018 1ohn Wiley & Sons, Inc All Rights Reserved. A Division of 2ohn Wiley & Sons Jas 618 2018-09-19 1

Explanation / Answer

Part A.

Time taken by ball to reach the net will be given by:

distance = Velocity*time

Since there is no acceleration in horizontal direction, So Using above equation

Given that net is at 12 m and 0.900 m high

Now,

12 = V0x*t

t = 12/(V0*cos theta)

theta = 0 deg, since ball leaves racquet horizontally

t = 12/(29.5*cos 0) = 0.407 sec

Now at this time height of ball will be, Using

y - y0 = V0y*t + 0.5*a*t^2

y0 = 2.38 m

V0y = 0

t = 0.407 sec

a = -g = -9.81 m/sec^2

So,

y = 2.38 + 0*0.407 - 0.5*9.81*0.407^2

y = 1.57 m

Since height of ball is higher than net so it will clear the net and distance between net and ball will be

d = 1.57 - 0.900 = 0.67 m

Part B.

When ball is struck at 5 deg below horizontal

Using same procedure to find time

12 = V0x*t

t = 12/(V0*cos theta)

theta = 5 deg, since ball leaves racquet horizontally

t = 12/(29.5*cos 5 deg) = 0.408 sec

Now at this time height of ball will be, Using

y - y0 = V0y*t + 0.5*a*t^2

y0 = 2.38 m

V0y = -29.5*sin 5 deg = -2.57

t = 0.408 sec

a = -g = -9.81 m/sec^2

So,

y = 2.38 - 2.57*0.408 - 0.5*9.81*0.408^2

y = 0.514 m

Since height of ball is lower than net so it will not clear the net and distance between net and ball will be

d = 0.514 - 0.900 = -0.386 m

Please Upvote.

Comment below if you have any doubt.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.