this is vertical launch 6 trials 1: 169.0 cm 2: 167.5 cm 3: 173.0 cm 4: 175.0 cm

Question

this is vertical launch

6 trials

1: 169.0 cm

2: 167.5 cm

3: 173.0 cm

4: 175.0 cm

5: 176.0 cm

6: 178.0 cm

and the average = 1.7308 m

this is launching at an angle (18 degrees)

six trials

1: 372.2 cm

2: 365.5 cm

3: 364.0 cm

4: 369.0 cm

5: 370.8 cm

6: 369.2 cm

the average displacement = 3.6845 m

y = 0.99 m

angle = 18 degrees

please help to find the initial velocity ><!!!!!

and I need step-by-step so that I can understand how to calculate it.

Ps. the initial velocity should be the same! (or about)

Explanation / Answer

a ) at maximum height, velocity = 0
0 = u^2 - 2gHmax [v^2 = u^2 + 2as]
u = (2gHmax)            taking Hmax as given 1.7308 m

   = 5.8244 m/sec

b ) let initial velocity = u

horizonatl component = ucos18

avg.displacement = ucos18 * t (time of flight)

now time of flight = time to reach maximum height + time to fall from maximum height to ground

          now,time to reach maximum height  = usin18/9.8           [v = u +at]

time to fall from maximum height to ground = (2 *(Hmax +0.99) * 9.8     [s = ut + 1/2at^2]

                where Hmax = u^2/2g

hence time of flight  = u/9.8 + (2 *(u^2sin^2(18)/2g +0.99) / 9.8)

since, avg.displacement = ucos18 * t (time of flight)

  3.6845 = ucos18 * {usin18/9.8 + (2 *(u^2sin^2(18)/2g +0.99) / 9.8)}

solving it for u we get,

   you can solve this equation with the help of available softwares

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