Information given:
h_{o} = 1.00 m
h_{o} = 1.00 m
t_{1} = 1.00s
t_{2} = 5.00s
D = (d_{x2} - d_{x1}) = 50.0 m
From O to P takes 1 s;
From P to Q takes 4 s;
From Q to f takes 1 s
Time for entire journey = 6 s
(a) Range = d_{xf}
There is no horizontal force and hence no horizontal acceleration.
Horizontal distance coverd in the 4 seconds (from P to Q) = D = 50m
Distance covered in 1 second (V_{x}) = ^{50m}/_{4s} = 12.5 m^{-2}
--- Eq1
Using x = x _{o} + v_{ox}t + ½at^{2}
x = 0 + (12.5 m^{-2}) (6s) + ½(0) (6^{2})
Range = 75m.
(b) Velocity of stone at O = Vector sum of v_{ix} and v_{iy}
v_{ix} = (12.5 m^{-2}) from Eq1
Vetical distance travelled by the ball
Applying y = y_{o} + v_{oy}t + ½at^{2}--- Eq2
Now
y_{f} = y_{0} (h = final height of ball = initial height of ball)
So Eq2 becomes h = h + v_{oy}t + ½at^{2}--- Eq2
0 = v_{oy} (6s) + ½ (-9.81ms^{-2}) (6s) ^{2}
v_{oy} = 29.4 ms^{-1}
By Pythagoras:
v_{i}^{2} = √(v_{ix}^{2} + v_{iy}^{2})
v_{i}^{2} = √[(12.5 ms^{-1})^{2} + (29.4 ms^{-1})^{2}]
v_{i} = 31.9 ms^{-1}
tan θ_{i} = (v_{iy} / v_{ix})
tan θ_{f} = (29.4 ms^{-1}) / (12.5 ms^{-1})
θ_{f} = tan^{-1} [(29.4 ms^{-1}) / (12.5 ms^{-1})]
θ_{f} = 67.0°
v_{i} = 31.9 ms^{-1} at an angle of elevation of 67.0°
(c) Applying y = y_{o} + v_{oy}t + ½at^{2}
h_{w} = 1m _{} + (29.4 ms^{-1}) (1s) + ½(-9.81ms^{-2}) (1s)^{2}
h_{w} = 22.5 m. |