Question

Problem 2. (25 points) A golfer hits a golf ball with an initial speed of 30 m/s at an angle of 35° with the ground. The ball flies over the level ground and lands at the same height from which it was hit. a) How long does it take for the ball to reach its highest point? b) What is the range of the ball? c) Find the velocity of the ball just before it hits the ground. d) After the ball hits the ground it rolls a short distance and then rolls off the horizontal surface of a cliff which is 23 meters above the ground below. If the ball slows down to 12 m/s just before it goes off the cliff, how far from the base of the cliff does the ball land?

Answer 1

Part-I Uyu typ X Hamon man Vuoso Part-1 stemte ucoso @ Along vertical. uy = wing upward Zoo ag=g downward At thmase y =0 . Let time at Hmone is to then, Vj = ly tayta »o= usinog ta ta= us » $ = 30 sen 35° second ha = 1.756 seconal.] Time of flight = axta (from symmetry + ļ= 2x1.756 second. - Range = 4 tp = (ueeso) tp =3069535) le > Range = 301009350)x 2x1.756m >> TRange = 86.3m) Dot = axta ② about Hmone).

© In time tq=2 ta Along vertical Vy = lytauty = uy-gx2ta Vy= uy-2gx using = usino-zusino Muy = -usino Since there is no accederation along horizontal so, Valeucoso - V = uceso t – weeho h »V = 3000335° – 30 sen 350 ) 181= 5u²cost of using au [181= 30mis ) In and part 19 11 =12mis JoJo I ne ay=g downward ay=h=23m doconward - "ay= Uyut thayth

ni-23= 0+46-9.8) te 9.8 So, distance à = let. >> x= 10x (en Im x=26m)