Fram the wndow of a builking, a bal is tossed from a height yo acave the ground

Question

Fram the wndow of a builking, a bal is tossed from a height yo acave the ground wth an inicial velocty of 8.20 m's and angle of 21.0P below the honzontal. It strikes the ground 5.00 s later (a) If the base of the bulding is taken to be the angin of the coordinates, with upward the pastive y-direction, whst a fo/lowing as necessary: vi. Assume St units. Do not subet tute numenical val-es; use variables orly.J e the inbal coordinates of the ba? (Use the tb) with the pesitivw x-drectien chosen to be sut the windom, fins the x and y comporents of the instial velocity lc) Find the equations for the x- and pcemponen patson as functors of time (Use the nasenne ois necessary: yo and t. Id) Hom far herizontaly trom the burss of the building does the ball strike the ground (a) Find the he ght fram swhich the ball was thecemn. th How lang does it take tha bal to reach a peint 10.0 m bel or the lwvel of launchenag

Explanation / Answer

a) Vertical component of velocity = 8.2Sin 21= 2.94 m/s

a = 9.81 m/s^2; vi = 2.94 m/s; t = 5s; s = ?; vf = ?

s = vit + 1/2 at^2 = (2.94 x 5) + (1/2 x 9.81 x 25)

s = 137.19m = height of building.

x = 0, y = 137.19m

b) Horizontal component = 8.2Cos 21 = 7.66 m/s

vi, x = + 7.66 m/s

vi, y = - 2.94 m/s

c) velocity = distance /time

Distance = Vel x time

x = + 7.66 t m

y = - ( 2.94 + 9.81 t) m

d) At 5 sec:, x = 5 x 7.64 = 38.28 m

e) Height of drop = 137.19 m (found in a)

f) s = ut + 1/2 at^2

10 = 2.94 t + 4.905 t^2

4.905 t^2 + 2.94 t - 10 = 0

t = 1.16s

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