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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




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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