Phy 101 Chapter > © In A Car Lift Usec © A Hydraulic Lift © We Show How To © A 3
ID: 1628575 • Letter: P
Question
Phy 101 Chapter > © In A Car Lift Usec © A Hydraulic Lift © We Show How To © A 3.00-liter Alum © When An Ideal G © The Helicopter V × www.webassign.net/web/Student/Assignment-Responses/submit2dep-15943643 Use the worked example above to help you solve this problem. A stone is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 21.0 m/s, as in the figure. The point of release is h = 46.0 m above the ground. (a) How long does it take for the stone to hit the ground? Your response differs from the correct answer by more than 10%. Double check your calculations (b) Find the stone's speed at impact. m/s (c) Find the horizontal range of the stone. EXERCISE HINTS: GETTING STARTED I I'M STUCK! suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building (h = 46.0 m) as in the example above. If it strikes the ground 35.0 m away, find the following. (Hint: For part (a) use the equation for the x-displacement to eliminate vot from the equation for the y-displacement.) (a) the time of flight 3.57 The x coordinate as a function of time is x(t) = cos(39.0) t, so the initial speed is vo = 2Kcos 39.0t), where Ax-35.0 and r 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 (b) the initial speed 12.6 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 (c) the speed and angle of the velocity vector with respect to the horizontal at impact A6:39Explanation / Answer
solving first question
initial height=46 m
initial angle=30 degree
velocity=21 m/s
let the total time is t
initiall vertical component of velocity=vsin(30)=10.5 m/s
we know
S=ut+0.5at^2
-46=10.5t-4.9t^2
we got t=4.3 sec
(b)vertical velocity=10.5-(9.8*4.3)=-31.64 m/s
horizontal velocity=21cos(30)=18.2 m/s
net velocity=36.5 m/s
(c)range=21cos(30)*4.3=78.2 m