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Question Number 183039 by Mastermind last updated on 18/Dec/22

A Golfer practising on a range with an accelerated tree 4.9m above the fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20m/s. (Assume the acceleration due to gravity is 9.8m/s^2 and the effect of air resistance maybe ignored unless othewise stated 1) How long after the ball leaves the club will it land on the fairway? 2) What horizontal distance will the ball travel before striking the fairway? 3) What is the acceleration of the ball 0.5s after being hit? 4) Calculate the speed of the ball 0.8s after it leaves the club? M.m

$$\mathrm{A}\:\mathrm{Golfer}\:\mathrm{practising}\:\mathrm{on}\:\mathrm{a}\:\mathrm{range} \\ $$$$\mathrm{with}\:\mathrm{an}\:\mathrm{accelerated}\:\mathrm{tree}\:\mathrm{4}.\mathrm{9m}\:\mathrm{above} \\ $$$$\mathrm{the}\:\mathrm{fairway}\:\mathrm{is}\:\mathrm{able}\:\mathrm{to}\:\mathrm{strike}\:\mathrm{a}\:\mathrm{ball} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{leaves}\:\mathrm{the}\:\mathrm{club}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{horizontal}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{20m}/\mathrm{s}. \\ $$$$\left(\mathrm{Assume}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\right. \\ $$$$\mathrm{gravity}\:\mathrm{is}\:\mathrm{9}.\mathrm{8m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{effect}\:\mathrm{of}\:\mathrm{air} \\ $$$$\mathrm{resistance}\:\mathrm{maybe}\:\mathrm{ignored}\:\mathrm{unless} \\ $$$$\mathrm{othewise}\:\mathrm{stated}\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{How}\:\mathrm{long}\:\mathrm{after}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{leaves}\:\mathrm{the} \\ $$$$\mathrm{club}\:\mathrm{will}\:\mathrm{it}\:\mathrm{land}\:\mathrm{on}\:\mathrm{the}\:\mathrm{fairway}? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{What}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{will}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{travel}\:\mathrm{before}\:\mathrm{striking}\:\mathrm{the}\:\mathrm{fairway}? \\ $$$$\left.\mathrm{3}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball} \\ $$$$\mathrm{0}.\mathrm{5s}\:\mathrm{after}\:\mathrm{being}\:\mathrm{hit}? \\ $$$$\left.\mathrm{4}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\: \\ $$$$\mathrm{0}.\mathrm{8s}\:\mathrm{after}\:\mathrm{it}\:\mathrm{leaves}\:\mathrm{the}\:\mathrm{club}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Answered by mr W last updated on 19/Dec/22

1) 4.9=((9.8t^2 )/2) ⇒t=1 s 2) 20×1= 20m 3) at any time: a=9.8 m/s^2 4) v_h =20 m/s v_v =9.8×0.8=7.84 m/s v=(√(20^2 +7.84^2 ))≈21.5 m/s

$$\left.\mathrm{1}\right) \\ $$$$\mathrm{4}.\mathrm{9}=\frac{\mathrm{9}.\mathrm{8}{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{t}=\mathrm{1}\:{s} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{20}×\mathrm{1}=\:\mathrm{20}{m} \\ $$$$\left.\mathrm{3}\right) \\ $$$${at}\:{any}\:{time}:\:{a}=\mathrm{9}.\mathrm{8}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left.\mathrm{4}\right) \\ $$$${v}_{{h}} =\mathrm{20}\:{m}/{s} \\ $$$${v}_{{v}} =\mathrm{9}.\mathrm{8}×\mathrm{0}.\mathrm{8}=\mathrm{7}.\mathrm{84}\:{m}/{s} \\ $$$${v}=\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{7}.\mathrm{84}^{\mathrm{2}} }\approx\mathrm{21}.\mathrm{5}\:{m}/{s} \\ $$

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