Chapter 03, Problem 20 GO A golfer hits a shot to a green that is elevated 3.40

Question

Chapter 03, Problem 20 GO A golfer hits a shot to a green that is elevated 3.40 m above the point where the ball is struck. The ball leaves the dub at a speed of 19.9 m/s at an engle of 34.0 above the horizontal. It ses to its maximum height and then falls down to the green. Ignonng r resstance, find the speed of the balDust before it lands. the tolerance is +/-296 LENK T0 TEXT Question Attempts: Unlimited SAVE FOR LATER MapleNet All Rights Reserved A Division of Johney &.5ons.In ere to search /10/20127

Explanation / Answer

horizontal velocity = 19.9*cos(34 degrees) = 16.49 m/s

vertcal velocty V = 19.9*sin(34 degrees) = 11.12 m/s

max. height reached H = V^2/2g = (11.12)^2 / (2*9.8) = 6.30 m

elevation height = 3.40 m

=> remaining height = 6.30 - 3.40 = 2.9 m

speed at max. height = 0 m/s

so time taken to land s t

S = ut + 0.5*at^2

=> 2.9 = 0 + 0.5*9.8*t^2

=> t = 0.769 sec

so vertical velocity just before it lands

v = u + at = 0 + 9.8*0.769 = 7.53 m/s

horizontal velocity wll remain same = 16.49 m/s because of zero horizontal acceleration.

SO net velocity just before it lands

Vnet = [(16.49)^ + (7.53)^2]^0.5 = 18.12 m/s

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