Suppose the stone is thrown at an angle of 39.0° below the horizontal from the s

Question

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building (h = 47.0m) as in the example above. If it strikes the ground 46.3 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.) (a) the time of flight_____ s (b) the initial speed______ m/s (c) the speed and angle of the velocity vector with respect to the horizontal at impact

speed______ m/s

angle_______ ° below the horizontal

Please help this assignent is due tonight and I cant figure this out!!

Explanation / Answer

Let initial speed be U m/s and time of flight be T seconds.

Horizontal component velocity = Ucos = Ucos39
Initial Vertical component velcoity = Usin = Usin39

Height = 47 m and Range = 46.3 m .

X = Ucos39 x T
46.3 = Ucos39 x T

47 = ½(9.8)(T)² + Usin39xT

T = 46.3/Ucos39

47 = ½(9.8)(46.3/Ucos39)² + Usin39/(46.3/Ucos39)
47 = ½(9.8)(46.3/Ucos39)² + 46.3 Tan39 '
U = 42.7 m/s
a) T = 1.39s

Vertical componenet at impact = 42.7sin39 + (1.39)(9.8) = 40.4 m/s

b) Speed at impact = (42.7cos39)² + (40.4)² = 40.8 m/s

c) Angle =Arctan 40.4/42.7cos39 = 45 below horizontal

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