Question

A golf ball is hit with an initial speed of 50.0 m/s with a 7-iron. If the ball left at 40.0 degrees and landed at a

point 50.0m lower than the initial position:

a)How long was the ball in the air?

b)How faraway will the ball land?

c) What is the maximum height with respect to the landing point?

d)How long does it take for the ball to move ONLY a

distance , on the horizontal, away

sin(2e) R sin(20) g

from where it was hit?

Answer 1

E o somits. e ik and ithalateery Y-axis Given - the initial speed with which golf ball is hit is soms at an angle 4ől as shown'- - We will resolve .. the initial ... velocity in .. components & Usa ucosho .. Un Calong X-asis) Lamm 50x10syê =38-302 m/s. Earths Surface B ... and acceleration is towards earth surface in vertically downward direction since the ball experiences of rawitational force in ..... ... downwand direction here upward direction is taken the and dowald direction is taken -ve... . Chei-- 9:81. ml.s .. Also the vertical depth of landing kt. (B) is ..50.m.below leto. las Shalon Mencé in brojectile motion we can treat ! G - straight line ... mediomalong X-4 y axes independently

Ans. alony Y-axes...... - Usinuo = 50 sinuo =32-139 m/s -.-.9:21.m.152 also ... 5om displacement in -ve en 4 - direction is.. direction sce...... y ......S.... 50 mg e @ Apply 2 equation of motion.... ... let the time taken to go from A B ý t seconds ...sy. Vyb #iL qyt. =).-50 = 32-13.9 -4(9:81)ť - 4-got- 32013 g t - 50=0 Lance Et is a Quadratisciation in ... ter..... .. 10:3;2:1397# V6=32:139)277904903450) 214.90) 32-1395 V1039.91+980...... .. . 9.8... = 32-139 I 44.26 = taking only the - valve. 44:8737-857 seccand's go 8 ............ta ............................ ................

Ans. o let the height above pt o A be t... ... as shown in frig..... - Do In y direction from A c. at paging < Vy=ó only. No exis to there ... ply=0 Layanage then m e Vy = Um +2 Qy Sy ..... ..02(3,2-13.9) = 21.9.2.8.1.) H. ...Wy=32:13.9.m.........Ha 132-13922 – 52-64m . 2 (9.81)... Maximum height with respect to landing point ... hmarxa H'+50 m ... max = (52.64 +50) m = 102-64m

Ans. (d, Since speed in x-direction is uniform. So let the time required.. to trave R = 1² sinzo ? bet seconds .. ................................ ..................................... do we have R=Urt and I.x = Vo los 40 ...chere Vorud.co=480 te R = V. Sin20 ..7. (V6 (03.40°. 7Vo (2 Sino coso) = Vo sino oot = 2x 32.139 = 60 55 s 3o8
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