Information given:|
θ_{o} = 60.0°
v_{o}= 42.0 ms^{-1}
t_{1} = 5.50s
y_{o}= 0.00 m.
(a) Using the equation of motion in the vertical plane:
y = y_{o} + v_{oy}t + ½at^{2} --- Eq1
and v_{oy} = v_{o} sin 60.0° and a = g = -9.80 ms^{-2 }and t = 5.50s
So Eq 1 becomes: y = 0.00 + (42.0 ms^{-1} sin 60.0°) (5.50s) + ½ (-9.80 ms^{-2}) (5.50s)^{2}
y = 52.0 m
Hence h = 52.0 m
(b) Velocity of stone at A = Vector sum of v_{fx} and v_{fy}
Horizontal acceleration = 0.00 ms^{-2 }
Horizontal velocity: v_{fx} = v_{ox} + (0.00 ms^{-2}) (t)
v_{fx} = (v_{o}cos 60.0°) + (0.00^{} ms^{-2 })(5.50s)
v_{fx} = 21.0 ms^{-1}
Vertical acceleration = g ms^{-2 }
Vertical velocity: v_{y} = v_{oy} + (g) (t)
v_{fy} = (v_{o} sin 60.0°) + (-9.80 ms^{-2}) (5.50s)
v_{fy} = -17.5 ms^{-1}
By Pythagoras:
v_{f}^{2} = √(v_{fx}^{2} + v_{fy}^{2})
v_{f}^{2} = √[( 21.0 ms^{-1})^{2} + (-17.5 ms^{-1})^{2}]
v_{f} = 27.3 ms^{-1}
tan θ_{f} = (v_{y} / v_{x})
tan θ_{f} = (-17.5 ms^{-1}) / (21.0 ms^{-1})
θ_{f} = 39.8°
v_{o} = 27.3 ms^{-1} at and angle of depression of 39.8°
(c) At maximum height v_{y} = 0.00 ms^{-1}
From v_{y}^{2} = v_{oy}^{2} + 2gh
0_{}^{2} = [(42.0 ms^{-1}) sin 60.0°)]^{2} + 2 X (-9.80 ms^{-2}) H
H = 67.5 m. |