Learning Goal: To practice Problem-Solving Strategy 4.1 for gmiele mation proble

Question

Learning Goal: To practice Problem-Solving Strategy 4.1 for gmiele mation problems. Comect Its best to place the ornigin of the coordinate system at ground level below the launching point because inthis way all the paints of interes (the launching point and the landing poin:) will have postve ooordnatas. (Based on your "perenoa, you know that its gnerally easier to work with po"tva coordinate; ) Keep in mind, however, that thi; is an arbitrary choice oonger sclution of the problem wi nof depend on the location of the crigin of your coordinale syatem A rock thrown with spaod 6.50 m/s ard launch angia 30.0 (abave the horzontal) traveis a hornzontal datance of d 15.0 Im batora hiting the ground. From what height was the rock thrown? Use the value g 9.80o m/sor the troa-fal acceleraton Now. de ne mbols re reser ng mo and final portion. ve ocity and me. Your target varable s and smlar to the picture below . te intal y coordinate of the rock. Your po onal representa on should be complete now. Known Find Solve Part C Find the haighty, from which the rock waa launched Express your answer in meters to three significant figures. View Available Hintls)

Explanation / Answer

Please Do not downvote, as your picture is not very clear and unable to read the values of speed and distance in given question.

I'm assuming V = 6.50 m/sec and d = 15 m/sec

If these values are different let me know in the comments section, I will re-calculate.

Now Given that

Inital speed, V0 = 6.50 msec at 30 deg above horizontal

V0x = initial horizontal speed = V0*cos A = 6.50*cos 30 deg = 5.63 m/sec

V0y = initial Vertical speed = V0*sin A = 6.50*sin 30 deg = 3.25 m/sec

ax = horizontal acceleration = 0 m/sec^2

ay = Vertical acceleration = - g = -9.8 m/sec^2

d = horizontal range = 15 m (Again check this value and if different let me know)

Now Horizontal range is given by:

d = V0x*T

as horizontal velocity remains constant, So

T = d/V0x = 15/5.63 = 2.66 sec

T = time of projectile motion = 2.66 sec

Now Using 2nd kinematicequation in vertical direction

y = V0y*T + (1/2)*ay*T^2

y = 3.25*2.66 - 0.5*9.8*2.66^2

y = -26.0 m

Height of rock = 26.0 m

Again Please Do not downvote, without giving reason.

Please Upvote, If it works

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