Suppose the stone is thrown at an angle of 29.0 degree below the horizontal from

Question

Suppose the stone is thrown at an angle of 29.0 degree below the horizontal from the same building (h = 46.0 m) as in the example above. If it strikes the ground 74.0 m away, find the following. the time of flight 1.32 The x coordinate as a function of time is x(t) = cos(29.0)t, so the initial speed is v_0 = delta x/(cos 29.0 delta r), where delta x = 74.0 and delta t is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.06394436993246, since the stone is thrown down (and to the right). s the initial speed 64.25 You have a correct expression for the initial speed, but your answer to part (a) is wrong, so this answer is wrong too. m/s the speed and angle of the velocity vector with respect to the horizontal at impact 71.38 Speed This is the correct expression, but since your answer to part (a) is wrong, this answer is wrong too. m/s 38 angle What is the initial velocity? Is either component negative? Which component changes with time? How does it change? What equations govern the changes in the components of velocity? If you know the components of the final velocity, can you find its magnitude and direction? degree below the horizontal

Explanation / Answer

a) S=ut+0.5 a t^2

46=0+0.5*9.8*t^2

t= 3.06 s

b) Along horizontal

s= ut

74=u*3.06

u=24.18 m/s

c) Final vertical velocity= v=u+at

v= 9.8*3.06= 29.988 m/s

so speed= Sqrt (24.18^2 + 29.988^2) = 38.52 m/s

angle tan^-1 ( 29.988/24.18) = 51.12 degrees

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