x0 = 0, y0 = 27 Angle = 30 Initial speed = 54.36 we launch a ball twice: first o

Question

x0 = 0, y0 = 27

Angle = 30

Initial speed = 54.36

we launch a ball twice: first cm, second cm

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You now know the initial speed of the projectile and you will be told a starting position and assigned a firing angle. Calculate the range of your projectile, Show your work. At home you should calculate the maximum and minimum ranges based on the range of velocities estimated earlier. The ball will be returned to you for a test firing. You will be asked to launch your projectile at the assigned angle from the starting position Group Launch Angle = Predicted Average Range before launch.) ; (this must be shown to the instructor Minimum and Maximum Range = (based on deviation of Speed) Result of Launch = Were your results consistent with your prediction? Did your launch results fall within the deviation in ranges you estimated? 16

Explanation / Answer

The ball will first reach the highest point of its journey. Here, vy = 0

uy = 54.36sin30

so, using v = [u2 - 2gh]1/2

=> 0 = [(54.36sin30)2 - 2(9.8)h]1/2

=> h = 37.691 m

total height will now be: H = h + yo = 64.691 m

also, time taken to reach the highest point is: t = (uy - vy)/g = 2.773 s

now, H = u'yt' + (1/2)gt'2

But u' = v = 0 [at the highest point]

so, 64.691 = 0 + (1/2)(9.8)t'2

=> t' = 3.633s

therefore, T = 3.633 + 2.773 = 6.406 s

in this time, the range will be:

R = uxT = (54.36cos30)(6.406) = 301.60 m.

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